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I've got some homework that asks:

Determine the number of integer solutions of $x_1 + x_2 + x_3 + x_4 + x_5 = 32$ where $x_i > 3$ for $1 \le i \le 5$.

My solution is as follows:

$C(32 + 5 - 1, 32) = C(36, 32)$. This does not consider the restriction that $x_i > 3$.

If we remove $((3 + 1) \cdot 5) = 20$ possibilities from 32, we are left with 12 objects to distribute into 5 places. This becomes:

$C(12 + 5 - 1, 12) = C(16, 12) = 1820$ possible solutions.

..

Is my approach of "removing" the 4 possibilities from each of the 5 $x_i$'s the correct method? I think that this removes the possibilities of there being 0, 1, 2 or 3 objects in each of the 5 $x_i$.

Thanks

Tim
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    You could solve equation $x_1+\cdots+x_5=32-20,x_i\ge0$. – pointer May 16 '14 at 16:10
  • You've done fine, Tim, and your reasoning is sound (and your answer is indeed correct). – amWhy May 16 '14 at 16:31
  • How do you come up with the 20 possibilities? I assume that there is over counting if you count the number of solutions w/o the restriction of xi > 3 and therefore you to have subtract something. I'm not sure what that "something" is though. Can someone explain? – Extreme112 Jan 18 '15 at 02:19

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