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If eight different books, 5 math and 3 cs, are randomly arranged on a shelf, what's the probability that all 3 cs books are together?

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1

$begingroup$

Here's what I did, but I feel like it's wrong.

$$3C3 * 5!/8!$$

so that we're choosing all 3 cs books to be together, and then choosing the remaining 5 math books to be arranged on the shelf, and we divide by $8!$ since there are $8!$ ways of arranging all the books.

probability combinatorics

share|cite|improve this question

edited Mar 19 at 9:16

N. F. Taussig

45k103358

asked Mar 19 at 5:55

ZakuZaku

1729

$endgroup$

add a comment | 

1

$begingroup$

Here's what I did, but I feel like it's wrong.

$$3C3 * 5!/8!$$

so that we're choosing all 3 cs books to be together, and then choosing the remaining 5 math books to be arranged on the shelf, and we divide by $8!$ since there are $8!$ ways of arranging all the books.

probability combinatorics

share|cite|improve this question

edited Mar 19 at 9:16

N. F. Taussig

45k103358

asked Mar 19 at 5:55

ZakuZaku

1729

$endgroup$

add a comment | 

$begingroup$

Here's what I did, but I feel like it's wrong.

$$3C3 * 5!/8!$$

so that we're choosing all 3 cs books to be together, and then choosing the remaining 5 math books to be arranged on the shelf, and we divide by $8!$ since there are $8!$ ways of arranging all the books.

probability combinatorics

share|cite|improve this question

edited Mar 19 at 9:16

N. F. Taussig

45k103358

asked Mar 19 at 5:55

ZakuZaku

1729

$endgroup$

share|cite|improve this question

edited Mar 19 at 9:16

N. F. Taussig

45k103358

asked Mar 19 at 5:55

ZakuZaku

1729

share|cite|improve this question

edited Mar 19 at 9:16

N. F. Taussig

45k103358

asked Mar 19 at 5:55

ZakuZaku

1729

edited Mar 19 at 9:16

N. F. Taussig

45k103358

edited Mar 19 at 9:16

N. F. Taussig

45k103358

asked Mar 19 at 5:55

ZakuZaku

1729

asked Mar 19 at 5:55

ZakuZaku

1729

2 Answers
2

active

oldest

votes

4

$begingroup$

Your answer is wrong because there are not $binom 3 3$ possible locations for the CS books. You should see that as relatively obvious since $binom 3 3 = 1$ which obviously doesn't make sense.

---

Let's try a simpler approach. Imagine treating the $3$ CS books as a single book in the group of $8$ - which becomes $6$ books in light of this interpretation. By treating the three as a single book, this guarantees the three are together, making things much simpler!

How many ways can you arrange this group of $6$ books? That would be $6!$ ways.

Within the triplet of three books, you can arrange them in $3!$ further ways. We do care about the order, which is why we don't use combinations here. Since this lends us to having $3!$ permutations per permutation of the overall set with the triplet, we multiply $6!$ and $3!$.

Overall, there are $8!$ ways to arrange the set of books altogether. Thus, the probability the three end up together is given by

$$frac{text{desired arrangements}}{text{all possible arrangements}} = frac{6! cdot 3!}{8!} = frac{3}{28} approx 10.714%$$

share|cite|improve this answer

answered Mar 19 at 7:30

Eevee TrainerEevee Trainer

9,70031740

$endgroup$

add a comment | 

2

$begingroup$

As an alternative to the nice solution provided by Eevee Trainer, we take as our sample space the $binom{8}{3}$ ways of selecting three of the eight positions for the computer science books. If the three computer science books are placed consecutively, the leftmost book in the block of three computer science books must be placed in one of the first six positions. Hence, the probability that the three computer science books are placed consecutively is
$$frac{6}{binom{8}{3}} = frac{3}{28}$$

share|cite|improve this answer

answered Mar 19 at 10:55

N. F. TaussigN. F. Taussig

45k103358

$endgroup$

add a comment | 

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4

$begingroup$

Your answer is wrong because there are not $binom 3 3$ possible locations for the CS books. You should see that as relatively obvious since $binom 3 3 = 1$ which obviously doesn't make sense.

---

Let's try a simpler approach. Imagine treating the $3$ CS books as a single book in the group of $8$ - which becomes $6$ books in light of this interpretation. By treating the three as a single book, this guarantees the three are together, making things much simpler!

How many ways can you arrange this group of $6$ books? That would be $6!$ ways.

Within the triplet of three books, you can arrange them in $3!$ further ways. We do care about the order, which is why we don't use combinations here. Since this lends us to having $3!$ permutations per permutation of the overall set with the triplet, we multiply $6!$ and $3!$.

Overall, there are $8!$ ways to arrange the set of books altogether. Thus, the probability the three end up together is given by

$$frac{text{desired arrangements}}{text{all possible arrangements}} = frac{6! cdot 3!}{8!} = frac{3}{28} approx 10.714%$$

share|cite|improve this answer

answered Mar 19 at 7:30

Eevee TrainerEevee Trainer

9,70031740

$endgroup$

add a comment | 

4

$begingroup$

Your answer is wrong because there are not $binom 3 3$ possible locations for the CS books. You should see that as relatively obvious since $binom 3 3 = 1$ which obviously doesn't make sense.

---

Let's try a simpler approach. Imagine treating the $3$ CS books as a single book in the group of $8$ - which becomes $6$ books in light of this interpretation. By treating the three as a single book, this guarantees the three are together, making things much simpler!

How many ways can you arrange this group of $6$ books? That would be $6!$ ways.

Within the triplet of three books, you can arrange them in $3!$ further ways. We do care about the order, which is why we don't use combinations here. Since this lends us to having $3!$ permutations per permutation of the overall set with the triplet, we multiply $6!$ and $3!$.

Overall, there are $8!$ ways to arrange the set of books altogether. Thus, the probability the three end up together is given by

$$frac{text{desired arrangements}}{text{all possible arrangements}} = frac{6! cdot 3!}{8!} = frac{3}{28} approx 10.714%$$

share|cite|improve this answer

answered Mar 19 at 7:30

Eevee TrainerEevee Trainer

9,70031740

$endgroup$

add a comment | 

$begingroup$

Your answer is wrong because there are not $binom 3 3$ possible locations for the CS books. You should see that as relatively obvious since $binom 3 3 = 1$ which obviously doesn't make sense.

---

Let's try a simpler approach. Imagine treating the $3$ CS books as a single book in the group of $8$ - which becomes $6$ books in light of this interpretation. By treating the three as a single book, this guarantees the three are together, making things much simpler!

How many ways can you arrange this group of $6$ books? That would be $6!$ ways.

Within the triplet of three books, you can arrange them in $3!$ further ways. We do care about the order, which is why we don't use combinations here. Since this lends us to having $3!$ permutations per permutation of the overall set with the triplet, we multiply $6!$ and $3!$.

Overall, there are $8!$ ways to arrange the set of books altogether. Thus, the probability the three end up together is given by

$$frac{text{desired arrangements}}{text{all possible arrangements}} = frac{6! cdot 3!}{8!} = frac{3}{28} approx 10.714%$$

share|cite|improve this answer

answered Mar 19 at 7:30

Eevee TrainerEevee Trainer

9,70031740

$endgroup$

share|cite|improve this answer

answered Mar 19 at 7:30

Eevee TrainerEevee Trainer

9,70031740

answered Mar 19 at 7:30

Eevee TrainerEevee Trainer

9,70031740

answered Mar 19 at 7:30

Eevee TrainerEevee Trainer

9,70031740

2

$begingroup$

As an alternative to the nice solution provided by Eevee Trainer, we take as our sample space the $binom{8}{3}$ ways of selecting three of the eight positions for the computer science books. If the three computer science books are placed consecutively, the leftmost book in the block of three computer science books must be placed in one of the first six positions. Hence, the probability that the three computer science books are placed consecutively is
$$frac{6}{binom{8}{3}} = frac{3}{28}$$

share|cite|improve this answer

answered Mar 19 at 10:55

N. F. TaussigN. F. Taussig

45k103358

$endgroup$

add a comment | 

2

$begingroup$

As an alternative to the nice solution provided by Eevee Trainer, we take as our sample space the $binom{8}{3}$ ways of selecting three of the eight positions for the computer science books. If the three computer science books are placed consecutively, the leftmost book in the block of three computer science books must be placed in one of the first six positions. Hence, the probability that the three computer science books are placed consecutively is
$$frac{6}{binom{8}{3}} = frac{3}{28}$$

share|cite|improve this answer

answered Mar 19 at 10:55

N. F. TaussigN. F. Taussig

45k103358

$endgroup$

add a comment | 

$begingroup$

As an alternative to the nice solution provided by Eevee Trainer, we take as our sample space the $binom{8}{3}$ ways of selecting three of the eight positions for the computer science books. If the three computer science books are placed consecutively, the leftmost book in the block of three computer science books must be placed in one of the first six positions. Hence, the probability that the three computer science books are placed consecutively is
$$frac{6}{binom{8}{3}} = frac{3}{28}$$

share|cite|improve this answer

answered Mar 19 at 10:55

N. F. TaussigN. F. Taussig

45k103358

$endgroup$

share|cite|improve this answer

answered Mar 19 at 10:55

N. F. TaussigN. F. Taussig

45k103358

answered Mar 19 at 10:55

N. F. TaussigN. F. Taussig

45k103358

answered Mar 19 at 10:55

N. F. TaussigN. F. Taussig

45k103358