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There are $100$ cards with a unique number from $1$ to $100$ written over them. How many ways can someone pick exactly $5$ cards where the numbers on them sum to $100$?

I am not sure but this could be a variation of the subset sum problem. But one of my acquaintances got this problem on the GRE test, which seems unlikely given that the problem is beyond the mathematical skill level tested on the GRE. So I am asking is there a simplistic solution to this problem? If not, I am also happy with the non-simplistic solution with a bit of explanation, since I lack the mathematical aptitude required to understand the Wolfram Mathworld link.

5xum
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Shaown S.
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  • this is asking how many partitions there are of $100$ into exactly five non-zero parts. using the conjugate partition idea this problem is the same as that of finding five positive integers $a,b,c,d,e$ such that $a+2b+3c+4d+5e=100$ – David Holden Aug 07 '14 at 08:30
  • @DavidHolden Remember that they should be unique as well. – Arthur Aug 07 '14 at 08:45
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    http://math.stackexchange.com/questions/646705/counting-distinct-restricted-integer-partitions-of-n-into-exactly-k-distinct – Sidharth Ghoshal Aug 07 '14 at 08:55
  • @Arthur thx. my careless reading! – David Holden Aug 07 '14 at 09:31
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    The answer for this question is $[x^{100}z^{5}]\prod_{i=1}^{100}(1+x^{i}z)=25337.$ – nczksv May 16 '15 at 00:01
  • Thank you, @nczksv. But it would be really useful if could explain the logic behind your solution. It is also fine if you can direct me towards an explanation elsewhere on the web. Also I would prefer if you submit this as an answer so that if this checks out, I can accept this as an answer. – Shaown S. May 20 '15 at 09:42
  • The Comment by @nczksv refers to the coefficient of $x^{100}z^5$ in the expansion of polynomial $\prod_{i=1}^{100} (1+x^i z)$. The exponent of $z$ in any such expansion counts how many times the term $x^i z$ is used, rather than the term $1$ (since these are the only ways $z$ will appear in a sum of products). Those terms all have distinct exponents $i$ for $x$, and by inspection the $x^{100} z^5$ coefficient counts the number of ways five distinct exponents $i$ can add to $100$. – hardmath Jul 26 '15 at 15:39

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