x nCr 5 = 4[(x-1) nCr 4]
does anyone know how to solve for x here?
Keep in mind the following:
$n! = n\cdot (n-1)! = n(n-1)\cdot (n-2)!$
So, $n! = n(n-1)(n-2)\cdots (n-r+1)(n-r)!$
Then your equation becomes:
${_nC_r} = \dfrac{n!}{r!(n-r)!} = \dfrac{n(n-1)(n-2)\cdots (n-r+1) \cancel{(n-r)!}}{r!\cancel{(n-r)!}} = \dfrac{n(n-1)(n-2)\cdots (n-r+1)}{r!}$
Then see post #3.