# Thread: combinatorics: binomial coefficients

1. ## combinatorics: binomial coefficients

Hello.
I am looking for advice for the following question.

Find all values of n and k for which C(n, k+1) = 3*C(n, k)

Would I convert the equation into factorials using C(n, k) = n! / (k)!(n-k)!
Or is there something simpler that I am just not seeing

2. Originally Posted by chrisc
Hello.
I am looking for advice for the following question.

Find all values of n and k for which C(n, k+1) = 3*C(n, k)

Would I convert the equation into factorials using C(n, k) = n! / (k)!(n-k)!
Or is there something simpler that I am just not seeing
hi Chris,

You do have a choice here....

$\displaystyle \displaystyle\binom{n}{k+1}=3\binom{n}{k}\Rightarr ow\frac{n!}{(n-k-1)!(k+1)!}=3\frac{n!}{(n-k)!k!}$

$\displaystyle \displaystyle\Rightarrow\frac{n!}{(n-k-1)!(k+1)k!}=3\frac{n!}{(n-k)(n-k-1)!k!}$

which allows you to simplify and find the relationship between n and k.

Alternatively, you could use the following identity, though there is little point!!

$\displaystyle \displaystyle\binom{n}{k}+\binom{n}{k+1}=\binom{n+ 1}{k+1}$

$\displaystyle \displaystyle\binom{n}{k+1}=3\binom{n}{k}\Rightarr ow\ 3\binom{n}{k}=\binom{n+1}{k+1}-\binom{n}{k}$

$\displaystyle \displaystyle\ 4\binom{n}{k}=\binom{n+1}{k+1}$

which, simplified leads to the same relationship.