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For any non-empty finite set $A$ of real numbers, let $S(A)$ be the sum of the elements in $A$. There are exactly $61$ $3$-element subsets $A$ of $(1,$ ... $,23)$ with $S(A) = 36$. The total number of $3$-element subsets of $(1,$ ... $, 23)$ with $S(A) < 36$ is $N$. Find $\frac{(N+ 45)}{25}$.

What I Tried: For some reason, the question did not really make that sense to me. For example, I did not fully understand the $2$nd sentence, did it mean that the set $A$ is actually the set $(1,$ ... $, 23) ?$ Probably not as then $S(A) \neq 36$. But then is $A$ another different set? If yes, how would we even start solving the question? Also if it meant that there are $61$ $3$-element subsets of $A$ , why is the set $(1,$ ... $, 23)$ even necessary?

As long as I am having problem understanding the question, I will have problems on solving it too.

Can anyone help?

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    $\begingroup$ No, $A$ is an arbitrary $3$ element subset of $\{1,\ldots,23\}$ that satisfies the property that $S(A)=36$. $\endgroup$ Dec 25, 2020 at 16:51

1 Answer 1

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We will use a bijection to solve this problem. For every 3-subset $(a,b,c)$ link this 3-subset to the 3-subset $(24-a,24-b,24-c)$. This will make a bijection between the 3-subsets with sum less than 36 to the 3-subsets with sum greater than 36, hence the quantity of them are equal. So we just have to decrease the number of 3-subsets with sum exactly equal to 36 -which is 61- from the number of all 3-subsets and divide it by 2 to find N. Because the number of all 3-subsets is $\binom{23}{3}=1771$ and so $N=\frac{1771-61}{2}=\frac{1710}{2}=855$ so the value required in the end of the problem is 36.

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  • $\begingroup$ Beatiful(+1) never would think like that $\endgroup$ Dec 25, 2020 at 17:08
  • $\begingroup$ Take my advice. Whenever something seems unable to count, search for a bijection 😂 $\endgroup$
    – Aryan
    Dec 25, 2020 at 17:08
  • $\begingroup$ Thanks by the way i have edited your post ,you had a typo hope you are fine with edit $\endgroup$ Dec 25, 2020 at 17:10
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    $\begingroup$ Oh thanks didn't know the command 😅 $\endgroup$
    – Aryan
    Dec 25, 2020 at 17:11
  • $\begingroup$ Got it, thanks for the help as well. $\endgroup$
    – Anonymous
    Dec 25, 2020 at 17:14

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