Kenneth Rosen Edition 7 Exercise 6.5 Question 9 (Page No. 432)

Question

A bagel shop has onion bagels, poppy seed bagels, egg bagels, salty bagels, pumpernickel bagels, sesame seed bagels, raisin bagels, and plain bagels. How many ways are there to choose

• A.  six bagels?
• B.  a dozen bagels?
• C.  two dozen bagels?
• D.  a dozen bagels with at least one of each kind?
• E.  a dozen bagels with at least three egg bagels and no more than two salty bagels?

Comments

In order to select six bagels, we have

8 ways to choose first bagel,

8 ways to choose 2nd bagel

and so on…  = 8^6 ways.

Why this approach is wrong?

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@taran97

Product rule make assumptions that order is important. So, when you are doing 8^6 you are saying that choosing (Bagel1, Bagel2, Bagel3, Bagel4, Bagel5, Bagel6) is different than (Bagel2, Bagel1, Bagel3, Bagel4, Bagel5, Bagel6).

But this is not true in the above question as here it is mentioned number of ways to choose (not number of ways to select). So, (Bagel1, Bagel2, Bagel3, Bagel4, Bagel5, Bagel6) is same as (Bagel2, Bagel1, Bagel3, Bagel4, Bagel5, Bagel6).

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I think you're overcouting the same thing.

For example if I choose first 3 as salty and next as egg

Another is also possible with your formula

First I choose 3 egg and 3 salty as the order doens't matter they both are same.

For these kind of one's you're Couting for

6!/3!3! Times which has to be counted for only one time.

Answer 1

A).

There are 8 kinds of bagels (n = 8), and we have to choose 6 bagels (r = 6). In this problem, order doesn’t matter (combination) and repeats are okay (we assume that the shop won’t run out of any kind of bagels). Therefore, there are C(n + r − 1, r) = C(8 + 6 − 1, 6) = C(13, 6) = 13!/(7! × 6!) = 1716 ways to choose.

B).

There are 8 kinds of bagels (n = 8), and we have to choose 12 bagels (r = 12). In this problem, order doesn’t matter (combination) and repeats are okay (we assume that the shop won’t run out of any kind of bagels). Therefore, there are C(n + r − 1, r) = C(8 + 12 − 1, 12) = C(19, 12) = 19!/(12! × 7!) = 50, 388 ways to choose.

C).

There are 8 kinds of bagels (n = 8), and we have to choose 24 bagels (r = 24). In this problem, order doesn’t matter (combination) and repeats are okay (we assume that the shop won’t run out of any kind of bagels). Therefore, there are C(n + r − 1, r) = C(8 + 24− 1, 24) = C(31, 24) = 31!/(24! × 7!) = 2629575 ways to choose.

D).

There are 8 kinds of bagels (n = 8), and we have to choose 4 bagels (r = 4) (from 12 we selected 8 from each kind of bagels remaining  left is 12-8=4). In this problem, order doesn’t matter (combination) and repeats are okay (we assume that the shop won’t run out of any kind of bagels). Therefore, there are C(n + r − 1, r) = C(8 + 4 − 1, 4) = C(11, 4) = 11!/(4! × 7!) = 330 ways to choose.

E).

There are 8 kinds of bagels (n = 8), and we have to choose 9 bagels (r = 4) (12 − 3 preselected egg bagles). In this problem, order doesn’t matter (combination) and repeats are okay (we assume that the shop won’t run out of any kind of bagels). We have allocated three choices to the egg bagel, but we have to figure out how to choose the salty bagels. We will add the number of ways to choose when we take exactly zero, one,​​​​and two salty bagels

No salty bagels:

n = 7 (7 kinds of bagels excluding the salty bagel), r = 9 (12 − 3 egg bagels). Therefore, there are C(n+r−1, r) = C(7+9−1, 9) = C(15, 9) = 15!/(9!×6!) = 455 ways to choose.

One salty bagel:

n = 7 (7 kinds of bagels excluding the salty bagel), r = 8 (12−3 egg bagels −1 salty bagel). Therefore, there are C(n + r − 1, r) = C(7 + 8 − 1, 8) = C(14, 8) = 14!/(8! × 6!) = 3, 003 ways to choose.

Two salty bagels:

n = 7 (7 kinds of bagels excluding the salty bagel), r = 7 (12−3 egg bagels −2 salty bagels). Therefore, there are C(n + r − 1, r) = C(7 + 7 − 1, 7) = C(13, 7) = 13!/(7! × 6!) = 1, 716 ways to choose.

So, the answer to this problem is: C(15, 9) + C(14, 8) + C(13, 7) = 455 + 3, 003 + 1, 716 = 5, 174 ways.

 

Comments

Option e answer is 9724 i think you did wrong calculation at the end.

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in this why using c(8,6) is a wrong approach?