Four married couples attend a party. Each person shakes hands with every other person, except their own...
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Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes?
The Next CEO of Stack OverflowHandshakes in a partyhow many people are at the partycounting hands shakeGraph Theory number of handshakes of couplesHandshakes in a partyHow many mixed double pairs can be made from 7 married couples provided that no husband and wife plays in a same set?A complicated handshake problemFinding the number of combinations.Handshake problem, each person can have 5 handshakesNumber of handshakes - exclusion apporachHandshake/pigeonhole principle problem?
$begingroup$
Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes?
My book gave the answer as $24$. I do not understand why.
I thought of it like this:
You have four pairs of couples, so you can think of it as
M1W2, M2W2, M3W3, M4W4,
where
M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have $4times 6$ handshakes, but in my answer, you are double counting.
How do I approach this problem?
combinatorics
edited Mar 18 at 22:15
Jack
27.6k1782203
asked Mar 17 at 4:41
ZakuZaku
1679
$endgroup$
|
show 3 more comments
$begingroup$
Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes?
My book gave the answer as $24$. I do not understand why.
I thought of it like this:
You have four pairs of couples, so you can think of it as
M1W2, M2W2, M3W3, M4W4,
where
M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have $4times 6$ handshakes, but in my answer, you are double counting.
How do I approach this problem?
combinatorics
edited Mar 18 at 22:15
Jack
27.6k1782203
asked Mar 17 at 4:41
ZakuZaku
1679
$endgroup$
3
$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
Mar 17 at 4:49
2
$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
Mar 17 at 4:56
4
$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
Mar 17 at 5:42
2
$begingroup$
Possible duplicate of Handshakes in a party
$endgroup$
– Xander Henderson
Mar 17 at 20:45
2
$begingroup$
@user21820 Hm, if it gets reopened, I'll post an answer. I don't think I see why it got closed. Sure it's an elementary problem, but it clearly shows effort and at least a part of the question is why the specific method used seems to be wrong but gives the correct answer.
$endgroup$
– M. Vinay
Mar 19 at 5:29
|
show 3 more comments
$begingroup$
Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes?
My book gave the answer as $24$. I do not understand why.
I thought of it like this:
You have four pairs of couples, so you can think of it as
M1W2, M2W2, M3W3, M4W4,
where
M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have $4times 6$ handshakes, but in my answer, you are double counting.
How do I approach this problem?
combinatorics
edited Mar 18 at 22:15
Jack
27.6k1782203
asked Mar 17 at 4:41
ZakuZaku
1679
$endgroup$
Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes?
My book gave the answer as $24$. I do not understand why.
I thought of it like this:
You have four pairs of couples, so you can think of it as
M1W2, M2W2, M3W3, M4W4,
where
M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have $4times 6$ handshakes, but in my answer, you are double counting.
How do I approach this problem?
combinatorics
combinatorics
edited Mar 18 at 22:15
Jack
27.6k1782203
asked Mar 17 at 4:41
ZakuZaku
1679
edited Mar 18 at 22:15
Jack
27.6k1782203
asked Mar 17 at 4:41
ZakuZaku
1679
edited Mar 18 at 22:15
Jack
27.6k1782203
edited Mar 18 at 22:15
Jack
27.6k1782203
edited Mar 18 at 22:15
Jack
27.6k1782203
27.6k1782203
asked Mar 17 at 4:41
ZakuZaku
1679
asked Mar 17 at 4:41
ZakuZaku
1679
asked Mar 17 at 4:41
ZakuZaku
1679
1679
3
$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
Mar 17 at 4:49
2
$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
Mar 17 at 4:56
4
$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
Mar 17 at 5:42
2
$begingroup$
Possible duplicate of Handshakes in a party
$endgroup$
– Xander Henderson
Mar 17 at 20:45
2
$begingroup$
@user21820 Hm, if it gets reopened, I'll post an answer. I don't think I see why it got closed. Sure it's an elementary problem, but it clearly shows effort and at least a part of the question is why the specific method used seems to be wrong but gives the correct answer.
$endgroup$
– M. Vinay
Mar 19 at 5:29
|
show 3 more comments
3
$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
Mar 17 at 4:49
2
$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
Mar 17 at 4:56
4
$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
Mar 17 at 5:42
2
$begingroup$
Possible duplicate of Handshakes in a party
$endgroup$
– Xander Henderson
Mar 17 at 20:45
2
$begingroup$
@user21820 Hm, if it gets reopened, I'll post an answer. I don't think I see why it got closed. Sure it's an elementary problem, but it clearly shows effort and at least a part of the question is why the specific method used seems to be wrong but gives the correct answer.
$endgroup$
– M. Vinay
Mar 19 at 5:29
3
3
$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
Mar 17 at 4:49
$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
Mar 17 at 4:49
2
2
$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
Mar 17 at 4:56
$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
Mar 17 at 4:56
4
4
$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
Mar 17 at 5:42
$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
Mar 17 at 5:42
2
2
$begingroup$
Possible duplicate of Handshakes in a party
$endgroup$
– Xander Henderson
Mar 17 at 20:45
$begingroup$
Possible duplicate of Handshakes in a party
$endgroup$
– Xander Henderson
Mar 17 at 20:45
2
2
$begingroup$
@user21820 Hm, if it gets reopened, I'll post an answer. I don't think I see why it got closed. Sure it's an elementary problem, but it clearly shows effort and at least a part of the question is why the specific method used seems to be wrong but gives the correct answer.
$endgroup$
– M. Vinay
Mar 19 at 5:29
$begingroup$
@user21820 Hm, if it gets reopened, I'll post an answer. I don't think I see why it got closed. Sure it's an elementary problem, but it clearly shows effort and at least a part of the question is why the specific method used seems to be wrong but gives the correct answer.
$endgroup$
– M. Vinay
Mar 19 at 5:29
|
show 3 more comments
8 Answers
8
active
oldest
votes
$begingroup$
$8$ people. Each experiences handshakes with $6$ people. There are $6times 8=48$ experiences of handshakes. Each handshake is experienced by two people so there $48$ experiences means $48div 2=24$ handshakes.
answered Mar 17 at 6:51
fleabloodfleablood
73.6k22891
$endgroup$
add a comment |
$begingroup$
Suppose the spouses were allowed to shake each other's hands. That would give you $binom{8}{2} = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.
answered Mar 17 at 5:02
Austin MohrAustin Mohr
20.8k35299
$endgroup$
$begingroup$
This uses Inclusion-Exclusion Principle.
$endgroup$
– smci
Mar 17 at 11:48
$begingroup$
Inclusion-Exclusion helps to find the cardinality of a union of non-disjoint sets. I'm merely using the fact that a set together with its complement (which are disjoint) comprise the entire universe.
$endgroup$
– Austin Mohr
Mar 18 at 2:30
$begingroup$
and that's just a case of Inclusion-Exclusion Principle. (By the way, the set we're enumerating here isn't the 'entire universe', since it's not the total number of handshakes, or handshakes with all people in the world, or even n-way handshakes with all people.)
$endgroup$
– smci
Mar 19 at 0:27
add a comment |
$begingroup$
You may proceed as follows using combinations:
- Number of all possible handshakes among 8 people: $color{blue}{binom{8}{2}}$
- Number of pairs who do not shake hands: $color{blue}{4}$
It follows:
$$mbox{number of hand shakes without pairs} = color{blue}{binom{8}{2}} - color{blue}{4} = frac{8cdot 7}{2} - 4 = 24$$
answered Mar 17 at 5:03
trancelocationtrancelocation
13.4k1827
$endgroup$
add a comment |
$begingroup$
Let's look at it not from individuals, but from couples. There are four couples, i.e. $3!=6$ meetings of couples. Per meeting of couples, there are four handshakes. This makes it $6times4=24$ handshakes.
Thanks @CJ Dennis for pointing out an error in the reasoning: It should, of course, be the sum, not the product, so the correct number of meetings of couples is
$sum_{k=1}^{n-1}k=frac{n(n-1)}{2}$.
answered Mar 17 at 14:47
dodidodi
693
$endgroup$
add a comment |
$begingroup$
Each line is a handshake between the required two people. There are 24 lines:
answered Mar 17 at 19:58
Witness Protection ID 44583292Witness Protection ID 44583292
23113
$endgroup$
add a comment |
$begingroup$
$k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:
$$sum_{i=1}^k (4k-4i) = sum_{i=1}^k4k - sum_{i=1}^k4i = 4k^2 - 4frac{k(k+1)}{2} = 4(k^2 - frac{k^2+k}{2}) = 4(k^2 - (frac{k^2}{2} + frac{k}{2})) = 4(frac{k^2}{2}-frac{k}{2}) = 2(k^2-k)$$
for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.
answered Mar 17 at 4:50
beefstew2011beefstew2011
687
$endgroup$
1
$begingroup$
Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
Mar 17 at 5:36
add a comment |
$begingroup$
A simple approach:
There are 8 person in total.
Each one will shake hands with 6 others.
Total shakehands from individual perspective: 6*8 gives 48
Actual shakehands: 48/2 = 24
answered Mar 17 at 8:10
Vijendra ParasharVijendra Parashar
191
$endgroup$
5
$begingroup$
How is different from fleablood's answer?
$endgroup$
– Toby Mak
Mar 17 at 8:46
1
$begingroup$
@TobyMak sorry, I really didn't see it. When I posted there were only four answers including mine. That answer was really not there, completely surprised. I don't know how this happened?
$endgroup$
– Vijendra Parashar
Mar 17 at 15:28
1
$begingroup$
I see. Since you wrote your answer independently from fleablood, it's only fair to keep your answer.
$endgroup$
– Toby Mak
Mar 18 at 8:22
add a comment |
$begingroup$
If all of them handshakes each other then there are 8!/2! =28 handshakes, but none of them handshake with their own spouse so their are 28-4=24 handshakes.
answered Mar 17 at 13:19
Chand16Chand16
276
$endgroup$
add a comment |
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8 Answers
8
active
oldest
votes
8 Answers
8
active
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$begingroup$
$8$ people. Each experiences handshakes with $6$ people. There are $6times 8=48$ experiences of handshakes. Each handshake is experienced by two people so there $48$ experiences means $48div 2=24$ handshakes.
answered Mar 17 at 6:51
fleabloodfleablood
73.6k22891
$endgroup$
add a comment |
$begingroup$
$8$ people. Each experiences handshakes with $6$ people. There are $6times 8=48$ experiences of handshakes. Each handshake is experienced by two people so there $48$ experiences means $48div 2=24$ handshakes.
answered Mar 17 at 6:51
fleabloodfleablood
73.6k22891
$endgroup$
add a comment |
$begingroup$
$8$ people. Each experiences handshakes with $6$ people. There are $6times 8=48$ experiences of handshakes. Each handshake is experienced by two people so there $48$ experiences means $48div 2=24$ handshakes.
answered Mar 17 at 6:51
fleabloodfleablood
73.6k22891
$endgroup$
$8$ people. Each experiences handshakes with $6$ people. There are $6times 8=48$ experiences of handshakes. Each handshake is experienced by two people so there $48$ experiences means $48div 2=24$ handshakes.
answered Mar 17 at 6:51
fleabloodfleablood
73.6k22891
answered Mar 17 at 6:51
fleabloodfleablood
73.6k22891
answered Mar 17 at 6:51
fleabloodfleablood
73.6k22891
answered Mar 17 at 6:51
fleabloodfleablood
73.6k22891
73.6k22891
add a comment |
add a comment |
$begingroup$
Suppose the spouses were allowed to shake each other's hands. That would give you $binom{8}{2} = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.
answered Mar 17 at 5:02
Austin MohrAustin Mohr
20.8k35299
$endgroup$
$begingroup$
This uses Inclusion-Exclusion Principle.
$endgroup$
– smci
Mar 17 at 11:48
$begingroup$
Inclusion-Exclusion helps to find the cardinality of a union of non-disjoint sets. I'm merely using the fact that a set together with its complement (which are disjoint) comprise the entire universe.
$endgroup$
– Austin Mohr
Mar 18 at 2:30
$begingroup$
and that's just a case of Inclusion-Exclusion Principle. (By the way, the set we're enumerating here isn't the 'entire universe', since it's not the total number of handshakes, or handshakes with all people in the world, or even n-way handshakes with all people.)
$endgroup$
– smci
Mar 19 at 0:27
add a comment |
$begingroup$
Suppose the spouses were allowed to shake each other's hands. That would give you $binom{8}{2} = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.
answered Mar 17 at 5:02
Austin MohrAustin Mohr
20.8k35299
$endgroup$
$begingroup$
This uses Inclusion-Exclusion Principle.
$endgroup$
– smci
Mar 17 at 11:48
$begingroup$
Inclusion-Exclusion helps to find the cardinality of a union of non-disjoint sets. I'm merely using the fact that a set together with its complement (which are disjoint) comprise the entire universe.
$endgroup$
– Austin Mohr
Mar 18 at 2:30
$begingroup$
and that's just a case of Inclusion-Exclusion Principle. (By the way, the set we're enumerating here isn't the 'entire universe', since it's not the total number of handshakes, or handshakes with all people in the world, or even n-way handshakes with all people.)
$endgroup$
– smci
Mar 19 at 0:27
add a comment |
$begingroup$
Suppose the spouses were allowed to shake each other's hands. That would give you $binom{8}{2} = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.
answered Mar 17 at 5:02
Austin MohrAustin Mohr
20.8k35299
$endgroup$
Suppose the spouses were allowed to shake each other's hands. That would give you $binom{8}{2} = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.
answered Mar 17 at 5:02
Austin MohrAustin Mohr
20.8k35299
answered Mar 17 at 5:02
Austin MohrAustin Mohr
20.8k35299
answered Mar 17 at 5:02
Austin MohrAustin Mohr
20.8k35299
answered Mar 17 at 5:02
Austin MohrAustin Mohr
20.8k35299
20.8k35299
$begingroup$
This uses Inclusion-Exclusion Principle.
$endgroup$
– smci
Mar 17 at 11:48
$begingroup$
Inclusion-Exclusion helps to find the cardinality of a union of non-disjoint sets. I'm merely using the fact that a set together with its complement (which are disjoint) comprise the entire universe.
$endgroup$
– Austin Mohr
Mar 18 at 2:30
$begingroup$
and that's just a case of Inclusion-Exclusion Principle. (By the way, the set we're enumerating here isn't the 'entire universe', since it's not the total number of handshakes, or handshakes with all people in the world, or even n-way handshakes with all people.)
$endgroup$
– smci
Mar 19 at 0:27
add a comment |
$begingroup$
This uses Inclusion-Exclusion Principle.
$endgroup$
– smci
Mar 17 at 11:48
$begingroup$
Inclusion-Exclusion helps to find the cardinality of a union of non-disjoint sets. I'm merely using the fact that a set together with its complement (which are disjoint) comprise the entire universe.
$endgroup$
– Austin Mohr
Mar 18 at 2:30
$begingroup$
and that's just a case of Inclusion-Exclusion Principle. (By the way, the set we're enumerating here isn't the 'entire universe', since it's not the total number of handshakes, or handshakes with all people in the world, or even n-way handshakes with all people.)
$endgroup$
– smci
Mar 19 at 0:27
$begingroup$
This uses Inclusion-Exclusion Principle.
$endgroup$
– smci
Mar 17 at 11:48
$begingroup$
This uses Inclusion-Exclusion Principle.
$endgroup$
– smci
Mar 17 at 11:48
$begingroup$
Inclusion-Exclusion helps to find the cardinality of a union of non-disjoint sets. I'm merely using the fact that a set together with its complement (which are disjoint) comprise the entire universe.
$endgroup$
– Austin Mohr
Mar 18 at 2:30
$begingroup$
Inclusion-Exclusion helps to find the cardinality of a union of non-disjoint sets. I'm merely using the fact that a set together with its complement (which are disjoint) comprise the entire universe.
$endgroup$
– Austin Mohr
Mar 18 at 2:30
$begingroup$
and that's just a case of Inclusion-Exclusion Principle. (By the way, the set we're enumerating here isn't the 'entire universe', since it's not the total number of handshakes, or handshakes with all people in the world, or even n-way handshakes with all people.)
$endgroup$
– smci
Mar 19 at 0:27
$begingroup$
and that's just a case of Inclusion-Exclusion Principle. (By the way, the set we're enumerating here isn't the 'entire universe', since it's not the total number of handshakes, or handshakes with all people in the world, or even n-way handshakes with all people.)
$endgroup$
– smci
Mar 19 at 0:27
add a comment |
$begingroup$
You may proceed as follows using combinations:
- Number of all possible handshakes among 8 people: $color{blue}{binom{8}{2}}$
- Number of pairs who do not shake hands: $color{blue}{4}$
It follows:
$$mbox{number of hand shakes without pairs} = color{blue}{binom{8}{2}} - color{blue}{4} = frac{8cdot 7}{2} - 4 = 24$$
answered Mar 17 at 5:03
trancelocationtrancelocation
13.4k1827
$endgroup$
add a comment |
$begingroup$
You may proceed as follows using combinations:
- Number of all possible handshakes among 8 people: $color{blue}{binom{8}{2}}$
- Number of pairs who do not shake hands: $color{blue}{4}$
It follows:
$$mbox{number of hand shakes without pairs} = color{blue}{binom{8}{2}} - color{blue}{4} = frac{8cdot 7}{2} - 4 = 24$$
answered Mar 17 at 5:03
trancelocationtrancelocation
13.4k1827
$endgroup$
add a comment |
$begingroup$
You may proceed as follows using combinations:
- Number of all possible handshakes among 8 people: $color{blue}{binom{8}{2}}$
- Number of pairs who do not shake hands: $color{blue}{4}$
It follows:
$$mbox{number of hand shakes without pairs} = color{blue}{binom{8}{2}} - color{blue}{4} = frac{8cdot 7}{2} - 4 = 24$$
answered Mar 17 at 5:03
trancelocationtrancelocation
13.4k1827
$endgroup$
You may proceed as follows using combinations:
- Number of all possible handshakes among 8 people: $color{blue}{binom{8}{2}}$
- Number of pairs who do not shake hands: $color{blue}{4}$
It follows:
$$mbox{number of hand shakes without pairs} = color{blue}{binom{8}{2}} - color{blue}{4} = frac{8cdot 7}{2} - 4 = 24$$
answered Mar 17 at 5:03
trancelocationtrancelocation
13.4k1827
answered Mar 17 at 5:03
trancelocationtrancelocation
13.4k1827
answered Mar 17 at 5:03
trancelocationtrancelocation
13.4k1827
answered Mar 17 at 5:03
trancelocationtrancelocation
13.4k1827
13.4k1827
add a comment |
add a comment |
$begingroup$
Let's look at it not from individuals, but from couples. There are four couples, i.e. $3!=6$ meetings of couples. Per meeting of couples, there are four handshakes. This makes it $6times4=24$ handshakes.
Thanks @CJ Dennis for pointing out an error in the reasoning: It should, of course, be the sum, not the product, so the correct number of meetings of couples is
$sum_{k=1}^{n-1}k=frac{n(n-1)}{2}$.
answered Mar 17 at 14:47
dodidodi
693
$endgroup$
add a comment |
$begingroup$
Let's look at it not from individuals, but from couples. There are four couples, i.e. $3!=6$ meetings of couples. Per meeting of couples, there are four handshakes. This makes it $6times4=24$ handshakes.
Thanks @CJ Dennis for pointing out an error in the reasoning: It should, of course, be the sum, not the product, so the correct number of meetings of couples is
$sum_{k=1}^{n-1}k=frac{n(n-1)}{2}$.
answered Mar 17 at 14:47
dodidodi
693
$endgroup$
add a comment |
$begingroup$
Let's look at it not from individuals, but from couples. There are four couples, i.e. $3!=6$ meetings of couples. Per meeting of couples, there are four handshakes. This makes it $6times4=24$ handshakes.
Thanks @CJ Dennis for pointing out an error in the reasoning: It should, of course, be the sum, not the product, so the correct number of meetings of couples is
$sum_{k=1}^{n-1}k=frac{n(n-1)}{2}$.
answered Mar 17 at 14:47
dodidodi
693
$endgroup$
Let's look at it not from individuals, but from couples. There are four couples, i.e. $3!=6$ meetings of couples. Per meeting of couples, there are four handshakes. This makes it $6times4=24$ handshakes.
Thanks @CJ Dennis for pointing out an error in the reasoning: It should, of course, be the sum, not the product, so the correct number of meetings of couples is
$sum_{k=1}^{n-1}k=frac{n(n-1)}{2}$.
answered Mar 17 at 14:47
dodidodi
693
edited Mar 18 at 9:53
answered Mar 17 at 14:47
dodidodi
693
answered Mar 17 at 14:47
dodidodi
693
answered Mar 17 at 14:47
dodidodi
693
693
add a comment |
add a comment |
$begingroup$
Each line is a handshake between the required two people. There are 24 lines:
answered Mar 17 at 19:58
Witness Protection ID 44583292Witness Protection ID 44583292
23113
$endgroup$
add a comment |
$begingroup$
Each line is a handshake between the required two people. There are 24 lines:
answered Mar 17 at 19:58
Witness Protection ID 44583292Witness Protection ID 44583292
23113
$endgroup$
add a comment |
$begingroup$
Each line is a handshake between the required two people. There are 24 lines:
answered Mar 17 at 19:58
Witness Protection ID 44583292Witness Protection ID 44583292
23113
$endgroup$
Each line is a handshake between the required two people. There are 24 lines:
answered Mar 17 at 19:58
Witness Protection ID 44583292Witness Protection ID 44583292
23113
answered Mar 17 at 19:58
Witness Protection ID 44583292Witness Protection ID 44583292
23113
answered Mar 17 at 19:58
Witness Protection ID 44583292Witness Protection ID 44583292
23113
answered Mar 17 at 19:58
Witness Protection ID 44583292Witness Protection ID 44583292
23113
23113
add a comment |
add a comment |
$begingroup$
$k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:
$$sum_{i=1}^k (4k-4i) = sum_{i=1}^k4k - sum_{i=1}^k4i = 4k^2 - 4frac{k(k+1)}{2} = 4(k^2 - frac{k^2+k}{2}) = 4(k^2 - (frac{k^2}{2} + frac{k}{2})) = 4(frac{k^2}{2}-frac{k}{2}) = 2(k^2-k)$$
for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.
answered Mar 17 at 4:50
beefstew2011beefstew2011
687
$endgroup$
1
$begingroup$
Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
Mar 17 at 5:36
add a comment |
$begingroup$
$k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:
$$sum_{i=1}^k (4k-4i) = sum_{i=1}^k4k - sum_{i=1}^k4i = 4k^2 - 4frac{k(k+1)}{2} = 4(k^2 - frac{k^2+k}{2}) = 4(k^2 - (frac{k^2}{2} + frac{k}{2})) = 4(frac{k^2}{2}-frac{k}{2}) = 2(k^2-k)$$
for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.
answered Mar 17 at 4:50
beefstew2011beefstew2011
687
$endgroup$
1
$begingroup$
Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
Mar 17 at 5:36
add a comment |
$begingroup$
$k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:
$$sum_{i=1}^k (4k-4i) = sum_{i=1}^k4k - sum_{i=1}^k4i = 4k^2 - 4frac{k(k+1)}{2} = 4(k^2 - frac{k^2+k}{2}) = 4(k^2 - (frac{k^2}{2} + frac{k}{2})) = 4(frac{k^2}{2}-frac{k}{2}) = 2(k^2-k)$$
for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.
answered Mar 17 at 4:50
beefstew2011beefstew2011
687
$endgroup$
$k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:
$$sum_{i=1}^k (4k-4i) = sum_{i=1}^k4k - sum_{i=1}^k4i = 4k^2 - 4frac{k(k+1)}{2} = 4(k^2 - frac{k^2+k}{2}) = 4(k^2 - (frac{k^2}{2} + frac{k}{2})) = 4(frac{k^2}{2}-frac{k}{2}) = 2(k^2-k)$$
for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.
answered Mar 17 at 4:50
beefstew2011beefstew2011
687
edited Mar 17 at 5:12
answered Mar 17 at 4:50
beefstew2011beefstew2011
687
answered Mar 17 at 4:50
beefstew2011beefstew2011
687
answered Mar 17 at 4:50
beefstew2011beefstew2011
687
687
1
$begingroup$
Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
Mar 17 at 5:36
add a comment |
1
$begingroup$
Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
Mar 17 at 5:36
1
1
$begingroup$
Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
Mar 17 at 5:36
$begingroup$
Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
Mar 17 at 5:36
add a comment |
$begingroup$
A simple approach:
There are 8 person in total.
Each one will shake hands with 6 others.
Total shakehands from individual perspective: 6*8 gives 48
Actual shakehands: 48/2 = 24
answered Mar 17 at 8:10
Vijendra ParasharVijendra Parashar
191
$endgroup$
5
$begingroup$
How is different from fleablood's answer?
$endgroup$
– Toby Mak
Mar 17 at 8:46
1
$begingroup$
@TobyMak sorry, I really didn't see it. When I posted there were only four answers including mine. That answer was really not there, completely surprised. I don't know how this happened?
$endgroup$
– Vijendra Parashar
Mar 17 at 15:28
1
$begingroup$
I see. Since you wrote your answer independently from fleablood, it's only fair to keep your answer.
$endgroup$
– Toby Mak
Mar 18 at 8:22
add a comment |
$begingroup$
A simple approach:
There are 8 person in total.
Each one will shake hands with 6 others.
Total shakehands from individual perspective: 6*8 gives 48
Actual shakehands: 48/2 = 24
answered Mar 17 at 8:10
Vijendra ParasharVijendra Parashar
191
$endgroup$
5
$begingroup$
How is different from fleablood's answer?
$endgroup$
– Toby Mak
Mar 17 at 8:46
1
$begingroup$
@TobyMak sorry, I really didn't see it. When I posted there were only four answers including mine. That answer was really not there, completely surprised. I don't know how this happened?
$endgroup$
– Vijendra Parashar
Mar 17 at 15:28
1
$begingroup$
I see. Since you wrote your answer independently from fleablood, it's only fair to keep your answer.
$endgroup$
– Toby Mak
Mar 18 at 8:22
add a comment |
$begingroup$
A simple approach:
There are 8 person in total.
Each one will shake hands with 6 others.
Total shakehands from individual perspective: 6*8 gives 48
Actual shakehands: 48/2 = 24
answered Mar 17 at 8:10
Vijendra ParasharVijendra Parashar
191
$endgroup$
A simple approach:
There are 8 person in total.
Each one will shake hands with 6 others.
Total shakehands from individual perspective: 6*8 gives 48
Actual shakehands: 48/2 = 24
answered Mar 17 at 8:10
Vijendra ParasharVijendra Parashar
191
answered Mar 17 at 8:10
Vijendra ParasharVijendra Parashar
191
answered Mar 17 at 8:10
Vijendra ParasharVijendra Parashar
191
answered Mar 17 at 8:10
Vijendra ParasharVijendra Parashar
191
191
5
$begingroup$
How is different from fleablood's answer?
$endgroup$
– Toby Mak
Mar 17 at 8:46
1
$begingroup$
@TobyMak sorry, I really didn't see it. When I posted there were only four answers including mine. That answer was really not there, completely surprised. I don't know how this happened?
$endgroup$
– Vijendra Parashar
Mar 17 at 15:28
1
$begingroup$
I see. Since you wrote your answer independently from fleablood, it's only fair to keep your answer.
$endgroup$
– Toby Mak
Mar 18 at 8:22
add a comment |
5
$begingroup$
How is different from fleablood's answer?
$endgroup$
– Toby Mak
Mar 17 at 8:46
1
$begingroup$
@TobyMak sorry, I really didn't see it. When I posted there were only four answers including mine. That answer was really not there, completely surprised. I don't know how this happened?
$endgroup$
– Vijendra Parashar
Mar 17 at 15:28
1
$begingroup$
I see. Since you wrote your answer independently from fleablood, it's only fair to keep your answer.
$endgroup$
– Toby Mak
Mar 18 at 8:22
5
5
$begingroup$
How is different from fleablood's answer?
$endgroup$
– Toby Mak
Mar 17 at 8:46
$begingroup$
How is different from fleablood's answer?
$endgroup$
– Toby Mak
Mar 17 at 8:46
1
1
$begingroup$
@TobyMak sorry, I really didn't see it. When I posted there were only four answers including mine. That answer was really not there, completely surprised. I don't know how this happened?
$endgroup$
– Vijendra Parashar
Mar 17 at 15:28
$begingroup$
@TobyMak sorry, I really didn't see it. When I posted there were only four answers including mine. That answer was really not there, completely surprised. I don't know how this happened?
$endgroup$
– Vijendra Parashar
Mar 17 at 15:28
1
1
$begingroup$
I see. Since you wrote your answer independently from fleablood, it's only fair to keep your answer.
$endgroup$
– Toby Mak
Mar 18 at 8:22
$begingroup$
I see. Since you wrote your answer independently from fleablood, it's only fair to keep your answer.
$endgroup$
– Toby Mak
Mar 18 at 8:22
add a comment |
$begingroup$
If all of them handshakes each other then there are 8!/2! =28 handshakes, but none of them handshake with their own spouse so their are 28-4=24 handshakes.
answered Mar 17 at 13:19
Chand16Chand16
276
$endgroup$
add a comment |
$begingroup$
If all of them handshakes each other then there are 8!/2! =28 handshakes, but none of them handshake with their own spouse so their are 28-4=24 handshakes.
answered Mar 17 at 13:19
Chand16Chand16
276
$endgroup$
add a comment |
$begingroup$
If all of them handshakes each other then there are 8!/2! =28 handshakes, but none of them handshake with their own spouse so their are 28-4=24 handshakes.
answered Mar 17 at 13:19
Chand16Chand16
276
$endgroup$
If all of them handshakes each other then there are 8!/2! =28 handshakes, but none of them handshake with their own spouse so their are 28-4=24 handshakes.
answered Mar 17 at 13:19
Chand16Chand16
276
answered Mar 17 at 13:19
Chand16Chand16
276
answered Mar 17 at 13:19
Chand16Chand16
276
answered Mar 17 at 13:19
Chand16Chand16
276
276
add a comment |
add a comment |
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3
$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
Mar 17 at 4:49
2
$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
Mar 17 at 4:56
4
$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
Mar 17 at 5:42
2
$begingroup$
Possible duplicate of Handshakes in a party
$endgroup$
– Xander Henderson
Mar 17 at 20:45
2
$begingroup$
@user21820 Hm, if it gets reopened, I'll post an answer. I don't think I see why it got closed. Sure it's an elementary problem, but it clearly shows effort and at least a part of the question is why the specific method used seems to be wrong but gives the correct answer.
$endgroup$
– M. Vinay
Mar 19 at 5:29