2. Suppose Tony has 20 donuts to distribute among his friends Cassie, Paul, Vict

Question

2. Suppose Tony has 20 donuts to distribute among his friends Cassie, Paul, Victoria, Ryan and Apara. But Tony might want to keep some donuts for himself (he also still might give them all away). The number of ways to distribute the donuts among Tony and his five friends is then the number of solutions to the inequality: where each xi (i - 1,2,3,4,5) is a nonnegative integer representing the number of donuts given to person (a) How many ways are there for Tony to distribute the donuts among himself and his 5 friends? b) Suppose Tony worked up a ravenous appetite doing a ton of Discrete Math, so he wants to keep at least 6 donuts for himself. How many ways are there for Tony to distribute the donuts?

Explanation / Answer

For an equation, x1 + x2 + x3 ... + xn = k, no of non negative integers = (k+n-1)C(n-1)

(a) x1 + x2 + x3 + x4 + x5 + x6 = 20

here n = 6, k = 20 [6 because Tony and 6 friends]

No of solution = (20+6-1)C(6-1) = 25C5 = 53130

(b) For keeping atleast 6 donuts with him, the equation becomes,

x1 + x2 + x3 + x4 + x5 + (x6 + 6) = 20

=> n=6, k =14

No of solution = (14+6-1)C(6-1) = 19C5 = 11628

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