MATH SOLVE

4 months ago

Q:
# A croissant 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?

Accepted Solution

A:

Answer:a) 110,880b) 8.61x10^37c) 7.18x10^19Step-by-step explanation:To find the amount of options, use the following combination equation. n!/[r! * (n - r)]In this equation, n represents the amount of possible options and r represents the amount being chosen at a time. We can now calculate out the answer for each possibility. a) In this case, n would equal 12 and r would equal 6. n!/[r! * (n - r)]12!/[6! * (12 - 6)]479001600/[720*6]479001600/4320110,880b) In this case, n would equal 36 and r would equal 6n!/[r! * (n - r)]36!/[6! * (12 - 6)]3.72x10^41/[720*6]3.72x10^41/43208.61x10^37c) For this one, n would equal 24 and r would equal 6. We would also then have to divide the answer at the end by 2. n!/[r! * (n - r)]12!/[6! * (12 - 6)]6.20x10^23/[720*6]6.20x10^23/43201.44x10^20 ÷ 2 = 7.18x10^19