The binomial coefficient

Calculating with factorials can be rough. Calculating with binomial coefficients can even be more rough. Two examples how you can counter the problems. Let’s look at two summations of binomial coefficients.

First: the sum \binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}

Let’s see

    \[\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}\]

We elaborate

    \[\frac{n!}{k!(n-k)!}+\frac{n!}{(k+1)!(n-(k+1))!}\]

Let’s create an equal denominator

    \[\frac{n!(k+1)}{(k+1)!(n-k)!}+\frac{n!(n-k)}{(k+1)!(n-k)!}\]

Simplify

    \[\frac{n!(k+1)+n!(n-k)}{(k+1)!(n-k)!}=\frac{n!(k+1+n-k)}{(k+1)!(n-k)!}\]

or

    \[\frac{n!(n+1)}{(k+1)!(n-k)!}=\frac{(n+1)!}{(k+1)!(n-k)!}=\binom{n+1}{k+1}\]

and there we are.

Now for the sum \binom{n}{k}+\binom{n}{k-1}=\binom{n+1}{k}

That is

    \[\binom{n}{k}+\binom{n}{k-1}=\binom{n+1}{k}\]

We tackle that in the same manner with

    \[\frac{n!}{k!(n-k)!}+\frac{n!}{(k-1)!(n-(k-1))!}\]

and equalizing the denominators

    \[\frac{n!(n-k+1)}{k!(n-k+1)!}+\frac{n!k}{k!(n-k+1)!}\]

so we can add the numerators:

    \[\frac{n!(n-k+1+k)}{k!(n-k+1)!}\]

which brings us to

    \[\frac{n!(n+1)}{k!(n-k+1)!}=\frac{(n+1)!}{k!(n+1-k)!}=\binom{n+1}{k}\]

Good boy! Down, down!