Answer to Question #228272 in Statistics and Probability for parth

Question #228272

A pub has 8 different beers on tap. On 4 different days each week they offer a discount price on one of the beers. The manager is deciding which beers to discount in a given week, suppose that they only want to choose the beers, not the specific day each is on special.

(a) If they want to select 4 different beers, how many different choices are there?

(b) Suppose that they now allow the possibility of selecting the same beer on multiple days, how many choices are there for selecting discounts for the week? This means they are choosing some number of beers, and the number of days each is on special, for a total of 4. For example one choice might be East End Lager for 2 days and Adelaide Bitter for 2 days.

(c) If they are still allowing the same beer to be chosen multiple times, but two of the beers are premium and they want premium beers to be on special on either 1 or 2 of the 4 days, then how many choices are there?


1
Expert's answer
2021-08-24T08:37:58-0400

(a)


"\\dbinom{8}{4}=\\dfrac{8!}{4!(8-4)!}=\\dfrac{8(7)(6)(5)}{1(2)(3)(4)}=70"

(b)


"(8)^4=4096"

(c)


"\\dbinom{2}{1}\\dbinom{6}{3}+\\dbinom{2}{2}\\dbinom{6}{2}"

"=2(\\dfrac{6!}{3!(6-3)!})+1(\\dfrac{6!}{2!(6-2)!})"

"=2(\\dfrac{6(5)(4)}{1(2)(3)})+\\dfrac{6(5)}{1(2)}=55"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS