Kids eating candy

Question

Determine how many ways I can distribute 80 candies to 3 kids, such that:

The first kid receives an arbitrary number of candies (possibly 0).
The second kid receives an even positive number of candies.
The third kid receives 0, 2, or 5 candies.
Every candy is distributed.

So far, I have gotten the generating function which is: (1+x+x^2+x^3...)(x^2+x^4+x^6...)(1+x^2+x^5).

score 1 · Accepted Answer · answered Jan 27 '19 at 23:29

1

Suppose the third kid receives $0$ candies. There are $40$ choices for the number of candies the second kid gets, and the number the third kid gets is then determined, so there are $40$ ways if the third kid get $0$ candies.

Do a similar thing if he gets $2$ or $5$ candies.

answered Jan 27 '19 at 23:29

saulspatz

53,131

What do you get? I tried 117 but that was wrong. – Math_Guy Jan 27 '19 at 23:41
I get $40+39+37=116$ – saulspatz Jan 28 '19 at 06:51
thank you for helping me – Math_Guy Jan 28 '19 at 21:33
@Math_Guy My pleasure – saulspatz Jan 28 '19 at 22:26

Kids eating candy

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