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?

Answer 1

Answer:

(a)262144

(b)68719476736  

(c)$4.72*10^{21}$

(d)4096

(e)343

Step-by-step explanation:

Since there are 8 kinds of bagels, at each bagels we have 8 different ways to pick a bagels. With n bagels there would be $8^n$ ways to choose bagels.

(a) there are $8^6 = 262144$ ways

(b) there are $8^{12} = 68719476736$ ways

(c) there are $8^{24} \approx 4.72*10^{21}$ways

(d) A dozen bagels with at least 1 of each kind would lock down the first 8 bagels to exactly be one of each kind, the rest (12 - 8 = 4 bagels) you can freely choose any of the 8

So there are $8^4 = 4096$ ways

(e) A dozen with at least 3 egg bagels and at most 2 salty bagels would lock down the first 5 bagels to be that. Furthermore, of the rest (8 - 5 = 3 bagels) you can only have 7 option each.

So there are$7^3 = 343$ ways