How many solutions for the equation $Y_1 + Y_2 + Y_3 + Y_4 + Y_5 = 6$?

Asked Jan 15 '17 at 05:07

Active Jan 15 '17 at 05:26

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How many solutions for the equation $Y_1 + Y_2 + Y_3 + Y_4 + Y_5 = 6$ ?

where $0\leq Y_1 \leq 3$ and $1\leq Y_2 \leq3$ and $Y_3,Y_4,Y_5\geq 0$ and $Y1,Y2,Y3,Y4,Y5 $ are integers.

I tried to apply brute force and got following equations :

Am I going right here ?

EDIT: Using generating functions gives $106$, but what is the error in above method.

edited Jan 15 '17 at 05:26

asked Jan 15 '17 at 05:07

Jon Garrick

Add that they are integers. – Gyanshu Jan 15 '17 at 05:11
1

From a program I put together, it appears the answer is 106 (if this helps). – Mark Schultz-Wu Jan 15 '17 at 05:21
@Mark Yes,I used generating functions and got 106. I'll edit that. But, why above method is not right ? – Jon Garrick Jan 15 '17 at 05:23
What you have noticed is that the number of ways to sum three numbers to make $n$ is $T_{n+1},$ the $(n+1)$st triangular number, which is $\frac 12(n+1)(n+2)$ – Ross Millikan Jan 15 '17 at 05:24

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