1. ## combinations equation.

x nCr 5 = 4[(x-1) nCr 4]

does anyone know how to solve for x here?

2. ## Re: combinations equation.

use

$$\binom {x} {5} = \binom {x - 1} {5} + \binom {x - 1} {4}$$

and simplify

3. ## Re: combinations equation.

Originally Posted by edwardkiely
x nCr 5 = 4[(x-1) nCr 4]

does anyone know how to solve for x here?
Use the definition for ${_n C_r} = \dfrac{n(n-1)(n-2)\cdots (n-r+1)}{r!}$

$\dfrac{x(x-1)(x-2)(x-3)(x-4)}{5!} = 4\dfrac{(x-1)(x-2)(x-3)(x-4)}{4!}$

Cancel out $(x-1)(x-2)(x-3)(x-4)$ on both sides, multiply both sides by $5!$, you wind up with $x=4\cdot 5 = 20$.

4. ## Re: combinations equation.

Or, obviously, x=1,2,3, or 4, so long as you are using the definition that ${_nC_r} = 0$ if $0\le n<r$

5. ## Re: combinations equation.

we are supposed to be able to solve it just using the the formula n!/r!*(n-r)!

do you know how i can solve just using this

6. ## Re: combinations equation.

Originally Posted by edwardkiely
we are supposed to be able to solve it just using the the formula n!/r!*(n-r)!

do you know how i can solve just using this
Keep in mind the following:
$n! = n\cdot (n-1)! = n(n-1)\cdot (n-2)!$

So, $n! = n(n-1)(n-2)\cdots (n-r+1)(n-r)!$

${_nC_r} = \dfrac{n!}{r!(n-r)!} = \dfrac{n(n-1)(n-2)\cdots (n-r+1) \cancel{(n-r)!}}{r!\cancel{(n-r)!}} = \dfrac{n(n-1)(n-2)\cdots (n-r+1)}{r!}$