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This is from a previous question paper for an entrance exam I am preparing for. https://www.allen.ac.in/apps/exam-2014/jee-advanced-2014/pdf/JEE-Main-Advanced-P-I-Maths-Paper-with-solution.pdf (Link for the paper. The question number is $56$).

I wanted to solve this question using star-bars method. My idea was that after finding number of all possible combinations (that is coming out to be $n-1\choose r-1$ = $20-1\choose 5-1$) I would divide the answer by $5!$ as every solution that allows for a

The Phenotype
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Arya
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    $(a,b,c,d,e)$ is distinct from what? – Pedro Jan 27 '18 at 16:57
  • The full question is somehow not visible. Can someone please edit it? – Arya Jan 27 '18 at 16:57
  • I understand that the question means that the tuples are unique. But it should be I think anytime the inequality is satisfied. – Arya Jan 27 '18 at 16:59
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    Come off it! This is trivial. You don't need any fancy techniques like stars and bars. There are no solutions if $a>2$ and not many for each of $a=1$ and $a=2$. Just count them! – almagest Jan 27 '18 at 17:06
  • Solve the unrestricted problem $a+b+c+d+e=20$ then half of the solutions will be for $a<b$ then half of those for $a<b<c$ and so on ... minus the case when $a=b=c=d=e$ – rtybase Jan 27 '18 at 17:13
  • @almagest true i can count them but I was hoping to learn where I was going wrong. After all the constant might not be so low the next time I face such a question. – Arya Jan 27 '18 at 17:18
  • @rtybase How can I guarantee only half of them will have b>a? – Arya Jan 27 '18 at 17:25
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    Something like this https://math.stackexchange.com/questions/2594154/find-generating-function-for-x-1x-2x-3x-4-n-while-a-x-1-ne-x-2-b-x-1/ – rtybase Jan 27 '18 at 17:28

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If $a<b<c<d<e$ it is trivial that $a,b,c,d,e$ are distinct numbers, so you can shorten the title. We may set $a=x_1$ and $b=a+x_2$, $c=b+x_3$, $d=c+x_4$, $e=d+x_5$, so we are looking for the solutions of $$ 5x_1+4x_2+3x_3+2x_4+x_5 = 20 $$ with $x_1,\ldots,x_5\in\mathbb{N}^+$. By setting $x_k=1+y_k$, we are looking for the solutions of $$ 5y_1+4y_2+3y_3+2y_4+y_5 = 5 $$ with $y_1,\ldots,y_5\in\mathbb{N}$. The number of such solutions is the coefficient of $z^5$ in the Taylor series at the origin of $$ \frac{1}{(1-z^5)(1-z^4)(1-z^3)(1-z^2)(1-z)} $$ i.e. the number of integer partitions of $5$, $p(5)=\color{red}{7}$.

Jack D'Aurizio
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