There are 3 students and 10 advisors. How many ways can each student be assigned an advisor if no advisor...
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There are 3 students and 10 advisors. How many ways can each student be assigned an advisor if no advisor advises all 3 students?
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If I understand this question correctly, should we break this question up into multiple cases where an advisor can advise two students, and then another advisor can advise another? After that, we add that to the case where each student has a different advisor. Would that mean:
$$(10 * 9 * 8) + (10 *9 *1) $$
We multiply by 1 since one advisor must be repeated.
combinatorics
asked Mar 18 at 2:11
ZakuZaku
1679
$endgroup$
add a comment |
$begingroup$
If I understand this question correctly, should we break this question up into multiple cases where an advisor can advise two students, and then another advisor can advise another? After that, we add that to the case where each student has a different advisor. Would that mean:
$$(10 * 9 * 8) + (10 *9 *1) $$
We multiply by 1 since one advisor must be repeated.
combinatorics
asked Mar 18 at 2:11
ZakuZaku
1679
$endgroup$
$begingroup$
Not quite. For the case in which exactly two students have the same advisor, you must choose which two of the three students are assigned to that advisor.
$endgroup$
– N. F. Taussig
Mar 18 at 2:33
add a comment |
$begingroup$
If I understand this question correctly, should we break this question up into multiple cases where an advisor can advise two students, and then another advisor can advise another? After that, we add that to the case where each student has a different advisor. Would that mean:
$$(10 * 9 * 8) + (10 *9 *1) $$
We multiply by 1 since one advisor must be repeated.
combinatorics
asked Mar 18 at 2:11
ZakuZaku
1679
$endgroup$
If I understand this question correctly, should we break this question up into multiple cases where an advisor can advise two students, and then another advisor can advise another? After that, we add that to the case where each student has a different advisor. Would that mean:
$$(10 * 9 * 8) + (10 *9 *1) $$
We multiply by 1 since one advisor must be repeated.
combinatorics
combinatorics
asked Mar 18 at 2:11
ZakuZaku
1679
asked Mar 18 at 2:11
ZakuZaku
1679
asked Mar 18 at 2:11
ZakuZaku
1679
asked Mar 18 at 2:11
ZakuZaku
1679
asked Mar 18 at 2:11
ZakuZaku
1679
1679
$begingroup$
Not quite. For the case in which exactly two students have the same advisor, you must choose which two of the three students are assigned to that advisor.
$endgroup$
– N. F. Taussig
Mar 18 at 2:33
add a comment |
$begingroup$
Not quite. For the case in which exactly two students have the same advisor, you must choose which two of the three students are assigned to that advisor.
$endgroup$
– N. F. Taussig
Mar 18 at 2:33
$begingroup$
Not quite. For the case in which exactly two students have the same advisor, you must choose which two of the three students are assigned to that advisor.
$endgroup$
– N. F. Taussig
Mar 18 at 2:33
$begingroup$
Not quite. For the case in which exactly two students have the same advisor, you must choose which two of the three students are assigned to that advisor.
$endgroup$
– N. F. Taussig
Mar 18 at 2:33
add a comment |
1 Answer
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$begingroup$
Count the total number of assignments, then subtract the number of invalid ones. Usually the way to go when counting things.
$$10^3 - 10$$
It's easy to count the number of ways all students can be assigned to the same advisor: it's just 10.
$endgroup$
add a comment |
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$begingroup$
Count the total number of assignments, then subtract the number of invalid ones. Usually the way to go when counting things.
$$10^3 - 10$$
It's easy to count the number of ways all students can be assigned to the same advisor: it's just 10.
$endgroup$
add a comment |
$begingroup$
Count the total number of assignments, then subtract the number of invalid ones. Usually the way to go when counting things.
$$10^3 - 10$$
It's easy to count the number of ways all students can be assigned to the same advisor: it's just 10.
$endgroup$
add a comment |
$begingroup$
Count the total number of assignments, then subtract the number of invalid ones. Usually the way to go when counting things.
$$10^3 - 10$$
It's easy to count the number of ways all students can be assigned to the same advisor: it's just 10.
$endgroup$
Count the total number of assignments, then subtract the number of invalid ones. Usually the way to go when counting things.
$$10^3 - 10$$
It's easy to count the number of ways all students can be assigned to the same advisor: it's just 10.
answered Mar 18 at 2:31
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$begingroup$
Not quite. For the case in which exactly two students have the same advisor, you must choose which two of the three students are assigned to that advisor.
$endgroup$
– N. F. Taussig
Mar 18 at 2:33