Combinatorics question.

Question

Jenna hosts a cookout with 7 of her friends. She has purchased 8 burgers, 8 steaks, and 5 pieces of chicken. If she wants to cook 16 food items in total (each person gets two items), including at least 2 burgers and 1 steak, how many possibilities are there to choose from what she purchased? Note: Do not distinguish between who gets what, only consider the 16 food items.

Answer 1

Assume all burgers steaks and chicken are equivalent... there isn’t one steak better than another that need to be considered a different item. Then the only thing we are looking for how many ways can you make b+ s + c =16 where b>=2 and s >= 1

B=#Burgers. S=#steaks c =# chicken

Tom is right you really don’t have to consider the 2 burger and 1 steak rule since no matter what you will always have at least 3 burgers and at least 3 steaks!! (Think about it, if you try to choose 16 things even if you use all the chicken and all the steak that’s only 13 things so you still have to have three burgers)

lets look at it straightforwardly

we only need to know the number if burgers and steaks because once we know that there is only once choice but chicken.. whatever it takes to make 16 things

The number of burgers has to be 3,4, 5,6 ,7, or 8

A)if b=3 then since only 5 pieces of chicken, s= 8. Count= 1

B) if b= 4 then s = 7 or 8

Count = 2

C) if b = 5 then s = 6 or 7 or 8

Count = 3

D ) if b= 6 then s = 5,6,7,or 8

Count = 4

E) if b = 7 then s = 4,5,6,7,8.

Count = 5

F) if b= 8 then s = 3,4,5,6,7 or 8

count = 6

thats all the choices

1+2+3+4+5+6 = 21

Answer 2

There are 21 items total. As she is selecting 16, she is not selecting 5. The two burgers and 1 steak condition is irrelevant as, if she is cooking 16 items, she has to cook 2 burgers and 1 steak.

Thus, not selecting 5 from 3 groups has C(7,2) ways - C(n + k - 1,k - 1) is a well-known result.

C(7,2) = 7*6/2 = 21