<PRE><B>Question 1202676</B>
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.</PRE>