# If the total number of $3$-element subsets of $(1,$ ... $, 23)$ with $S(A) < 36$ is $N$. Find $\frac{(N+ 45)}{25}$.

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?

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

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.