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The question:

There are $4$ parts in a test. The first three parts carry $10$ marks each and the fourth part carries $20$ marks. Assuming that the marks are integers, find the number of ways a candidate can get a score of $30$ (on $50$).

What I did:

I know how to solve such problems. We have to basically find the number of integer solutions to the equation $ a + b + c + d = 30 $ where $a, b, c \le 10$ and $d \le 20$. For this, we associate polynomial $(1 + t + t^2 + \ldots t^{10})^3(1 + t + t^2 \ldots + t^{20})$ with this equation and find out the coefficient of $t^{30}$. But in this case, that becomes tough. Like, I write the polynomial as:

$$\left( \dfrac{1 - t^{11}}{1-t}\right)^3\cdot \left(\dfrac{1-t^{21}}{1-t}\right) $$ $$=\left( 1 - t^{11} \right)^3 \cdot \left(1-t^{21}\right)\cdot\left(1-t\right)^{-4} $$

But this doesn't help. How to proceed further this way? Or, is there any better method available to solve this problem?

Marc van Leeuwen
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Parth Thakkar
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    It gets a little tedious, but you can always use stars-and-bars together with inclusion-exclusion, as in my solution to this problem. – Brian M. Scott Nov 02 '13 at 09:45
  • Here is a part of answer http://math.stackexchange.com/questions/524780/how-find-number-of-integer-solutions-x-1x-2-cdotsx-n-m?rq=1 – Thomas Nov 02 '13 at 10:05
  • You know the general term of the expansion of $(1-t)^4$. Expand the first part. We get a term of $1$, for which we need the coefficient of $t^{30}$ in the expansion of $(1-t)^{-4}$. There is a $-3t^{11}$ term, for which we need the coefficient of $t^{19}$. There is a $-t^{21}$, for which we need the coefficient of $t^9$. And finally there is a $3t^{22}$, for which we need the coefficient of $t^8$. Alternately, ask Wolfram Alpha for the product. – André Nicolas Nov 02 '13 at 10:11
  • I never thought of direct expansion. This was rather silly on my part. Thanks! And as for the alternative, it is, funnily enough, not allowed to be used in exams! :D. – Parth Thakkar Nov 03 '13 at 15:08

1 Answers1

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You can easily compute $(1-t^{11})^3(1-t^{21})$ even by hand, as it has only $8$ terms: $$(1-3t^{11}+3t^{22}-t^{33})(1-t^{21})=1-3t^{11}-t^{21}+3t^{22}+3t^{32}-t^{33}-3t^{43}+t^{54} $$ The coefficient of any $t^k$ in $(1-t)^{-4}$ is $\binom{k+3}3$. So the coefficient of $t^{30}$ in the product of the above sum with $(1-t)^{-4}$ can be computed term by term, and is $$ 1\times\binom{33}3-3\times\binom{33-11}3-1\times\binom{33-21}3+3\times\binom{33-22}3=1111. $$ Note that only the first four of the eight terms computed above were needed here, the oter having too high degree to begin with.

Marc van Leeuwen
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