Acroissant shop has plain croissants, cherry croissants, chocolate croissants, almond croissants, apple croissants, and broccoli croissants. how many ways are there to choose

a)a dozen croissants?

b)three dozen croissants?

c)two dozen croissants with at least two of each kind?

d)two dozen croissants with no more than two broccoli croissants?

e)two dozen croissants with at least five chocolate croissants and at least three almond croissants?

f)two dozen croissants with at least one plain croissant, at least two cherry croissants, at least three chocolate croissants, at least one almond croissant, at least two apple croissants, and no more than three broccoli croissants?

A. 6 188

B. 749 398

C. 6 188

D. 52 975

E. 20 349

F. 11 316

Explanation:

(a) The shop has 6 types of croissants of which a dozen(12) has to be selected

Therefore n=6, r=12

Repetition of croissants is permitted

And C(n+r-1, r)

C(6+12-1, 12) = C(17, 12) = 17!÷ 12!(17-12)! = 17!÷12! 5! =6 188

(b) The shop has 6 types of croissants of which three dozen(36) has to be selected

Therefore n=6, r=36

Repetition of croissants is permitted

And C(n+r-1, r)

C(6+36-1, 12) = C(41, 36) = 41!÷ 36!(41-36)! = 41!÷36! 5! = 749 398

(c) The shop has 6 types of croissants of which two dozen(24) has to be selected

Let us first select 2 of each kind which 12 croissants in total. Then we still need to select the remaining 12 croissants

Therefore n=6, r=12

Repetition of croissants is permitted

And C(n+r-1, r)

C(6+12-1, 12) = C(17, 12) = 17!÷ 12!(17-12)! = 17!÷12! 5! =6 188

(d) The shop has 5 types of croissants of which two dozen(24) has to be selected

Therefore n=5, r=24

Repetition of croissants is permitted

And C(n+r-1, r)

C(5+24-1, 24) = C(28, 24) = 28!÷ 24!(28-24)! = 28!÷24! 4! = 20 475

answer: 2x + 2 − 2 = 10 − 2; subtraction property of equality

step-by-step explanation: