Hi there the question states 2CR=25 I am really new to this so any help would be appreciated it states to solve for R.

I tried to solve but got no where what i tried was to expand with the formula of nCr and no luck :(

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- Jul 28th 2013, 12:27 AMx83Pre- calc HELP! Combinations
Hi there the question states 2CR=25 I am really new to this so any help would be appreciated it states to solve for R.

I tried to solve but got no where what i tried was to expand with the formula of nCr and no luck :( - Jul 28th 2013, 12:49 AMShadowKnight8702Re: Pre- calc HELP! Combinations
2 choose R = 25 is not possible.

- Jul 28th 2013, 12:50 AMx83Re: Pre- calc HELP! Combinations
Alright

But if it was, what is the process of solving one of these guys? - Jul 29th 2013, 02:50 AMemakarovRe: Pre- calc HELP! Combinations
- Jul 29th 2013, 03:08 AMPlatoRe: Pre- calc HELP! Combinations
Using the notation $\displaystyle _NC_k$ unless you are in an advanced mathematics class it is necessary that $\displaystyle N\ge k.$

If that is true then $\displaystyle _NC_k=M$ implies $\displaystyle \frac{N!}{k!(N-k)!}=M$.

Example: $\displaystyle _{N}C_2=105$ implies $\displaystyle \frac{N(N-1)}{2}=105$, solve for N.