SOLUTION: A box of Mrs. Donuts contains 8 honey dipped, 6 Bavarians, and 7 chocolate-filled donuts. How many ways can 5 donuts be selected to meet each condition? a. 3 Bavarian and 2 choc

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Question 1202676: A box of Mrs. Donuts contains 8 honey dipped, 6 Bavarians, and 7 chocolate-filled donuts. How many ways can 5 donuts be selected to meet each condition?
a. 3 Bavarian and 2 chocolate filled?
b. 2 Bavarian, 2 honey-dipped, and 1 chocolate filled?
c. exactly 1 honey dipped?
Found 2 solutions by ikleyn, mananth:
Answer by ikleyn(52834)   (Show Source): You can put this solution on YOUR website!
.
A box of Mrs. Donuts contains 8 honey dipped, 6 Bavarians, and 7 chocolate-filled donuts.
How many ways can 5 donuts be selected to meet each condition?
a. 3 Bavarian and 2 chocolate filled?
b. 2 Bavarian, 2 honey-dipped, and 1 chocolate filled?
c. exactly 1 honey dipped?
~~~~~~~~~~~~~~~~ 

Total donuts originally is  8 + 6 + 7 = 21.


(a)   = 20*21 = 420.    ANSWER


(b)   = 15*28*7 = 2940.    ANSWER


(c)   =  = 8*715 = 5720.    ANSWER

Solved.



Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
a.
3 Bavarian and 2 chocolate filled
Use combination formula, we get:
C(6,3) * C(7,2)
= (6! / (3! * (6-3)!) ) * (7! / (2! * (7-2)!) )
= 20 * 21 = 420
420 ways to select 3 Bavarian and 2 chocolate-filled donuts.
b.
2 Bavarian, 2 honey-dipped, and 1 chocolate-filled donut. Use the combination formula
C(6,2) * C(8,2) * C(7,1)
= (6! / (2! * (6-2)!) ) * (8! / (2! * (8-2)!) ) * (7! / (1! * (7-1)!) )
= 15 * 28 * 7 = 2940
2940 ways to select 2 Bavarian, 2 honey-dipped, and 1 chocolate-filled donut.
c.
1 honey-dipped donut, we choose 1 honey-dipped donut from 8
and 4 donuts from the remaining 13. we get
C(8,1) * C(13,4) = (8! / (1! * (8-1)!) ) * (13! / (4! * (13-4)!) ) = 8 * 715 = 5720
5720 ways to select 1 honey-dipped donut.
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