How many different 5-person teams can be formed from a group

The point here is the following:
Suppose we are told that there are 10 ways to choose \(x\) people out of 5. What is \(x\)? \(C^x_5=10\) --> \(\frac{5!}{x!(5-x)!}=10\) --> \(x!(5-x)!=12\) --> \(x=3\) or \(x=2\). So we cannot determine single numerical value of \(x\). Note that in some cases we'll be able to find \(x\), as there will be only one solution for it, but generally when we are told that there are \(n\) ways to choose \(x\) out of \(m\) there will be (in most cases) two solutions of \(x\) possible.

But if we are told that there are 10 ways to choose 2 out of \(x\), then there will be only one value of \(x\) possible --> \(C^2_x=10\) --> \(\frac{x!}{2!(x-2)!}=10\) --> \(\frac{x(x-1)}{2!}=10\) --> \(x(x-1)=20\) --> \(x=5\).

In our original question, statement (1) says that there are 126 ways to choose 5 out of \(x+2\) --> there will be only one value possible for \(x+2\), so we can find \(x\). Sufficient.

Just to show how it can be done: \(C^5_{(x+2)}=126\) --> \((x-2)(x-1)x(x+1)(x+2)=5!*126=120*126=(8*5*3)*(9*7*2)=5*6*7*8*9\) --> \(x=7\). Basically we have that the product of five consecutive integers (\((x-2)(x-1)x(x+1)(x+2)\)) equal to some number (\(5!*126\)) --> only one such sequence is possible, hence even though we have the equation of 5th degree it will have only one positive integer solution.

Statement (2) says that there are 56 ways to choose 3 out of \(x+1\) --> there will be only one value possible for \(x+1\), so we can find \(x\). Sufficient.

\(C^3_{(x+1)}=56\) --> \((x-1)x(x+1)=3!*56=6*7*8\) --> \(x=7\). Again we have that the product of three consecutive integers (\((x-1)x(x+1)\)) equal to some number (\(3!*56\)) --> only one such sequence is possible, hence even though we have the equation of 3rd degree it will have only one positive integer solution.

Hope it helps.