gmatJP wrote:
HI AbhayPrasanna.. Thanks for the reply...
I understood the first half but I dont follow your explanation of 'the only non-unique solution will be for the "r" in nCr eg. (8C3 = 8C5) but for n not equal to m, nCr can never be equal to mCr' ...
How do you know (x+2) C 5 = 126 is computable
praveenism wrote:
I agree with GmatJP...but I think we need to assume that whatever is given in the options is assumed to be true.So the x has to have an integral value.
But my concern is..what if the polynomial equation so formed of degree 5 have multiple solutions. In that case, the options fails to answer the situation.
@Abhayprasanna : The logic is correct and infact I also choose D going by the same logic.
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.
\(\frac{5!}{x!(5-x)!}=10\) --> \(x!(5-x)!=12\) --> \(x=3\) or \(x=2\).
In this how does x!(5-x)! give us x as 3 or 2? what is the calculation like?