Question 755607
36. If the number of subsets with 4 elements of a set A is equal to the number of
subsets with 5 elements of the set, then the number of subsets with 3 elements
of this set is:
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One solution would come from the case that if A has 3 or fewer elements, there
would be 0 subsets with 4 elements and 0 subsets with 5 elements, and 0 = 0. 
Let's see if there are any other solutions: 

If the set contains N elements, then the number of subsets it has with K
elements is C(N,K) = {{{N!/(K!(N-K)!))}}}

Suppose set A has N elements, where integer N > 3.  Then we have

C(N,4) = C(N,5) 

{{{N!/(4!(N-4)!)}}} = {{{N!/  (5!(N-5)!  )}}} 

Cross multiply

N!5!(N-5)! = N!4!(N-4)!

Divide both sides by N!

5!(N-5)! = 4!(N-4)!

Write out or indicate all the factors of all the factorial:

5·4·3·2·1(N-5)(N-4)(N-3)···1 =  4·3·2·1(N-4)(N-3)···1 

Divide both sides through by all the common factors, and we have:

5(N-5) = 1
 5N-25 = 1
    5N = 26
     N = {{{26/5}}} = {{{5&1/5}}}

That is not an integer, so there are no other cases.

So A has fewer than 4 elements.  If A has 3 elements, there is 1
subset, the improper subset A). If A has 2,1,or 0 elements, then 
the answer is 0.

Answer: The number of subsets with 3 elements is 0 or 1, or "None of the above".

Edwin</pre>