# Simplifying ${n\choose k} - {n-1 \choose k}$.

How would I simplify $${n\choose k} - {n-1 \choose k}$$

I've expanded them into the binomial coefficient form $$\frac{n!}{k!(n-k)!} - \frac{(n-1)!}{k!(n-1-k)!}$$

but that's about all I've got.

I'm having trouble with factorials. Any suggestions?

• Note that the denominator should have $(n-k)!$ not $(n-k!)$
– qwr
Commented Jun 30, 2020 at 4:51

To explain why your expression is equal to $${n - 1 \choose k - 1}$$ combinatorially, note that to choose $$k$$ out of $$n$$ objects entails one of two things:

1. You choose the first object and $$k - 1$$ objects from the remaining $$n - 1$$ objects. Clearly, there are $${n - 1 \choose k - 1}$$ ways to choose $$k$$ objects like this.

2. You don't choose the first object, meaning you must choose $$k$$ objects from the remaining $$n - 1$$ objects. Here there are $$n - 1 \choose k$$ ways to do this.

Hence, $${n - 1 \choose k} + {n - 1 \choose k - 1} = {n \choose k}$$ from which the equality in other answers follows.

• I love this explanation! Very intuitive. Thank you +1 Commented Jun 30, 2020 at 3:32

You can write it as $$\frac{(n-1)!}{k!(n-k-1)!}\left[\frac{n}{n-k}-1\right]=\frac{(n-1)!}{(k-1)!(n-k)!}=\binom{n-1}{k-1}$$

Use the fact that $$\binom n k = \binom {n - 1}{k - 1} + \binom {n - 1} k$$ for all integers $$1 \leq k \leq n - 1.$$

To expand on Anurag A's answer:

When working with combinatorial identities, $$n!$$ and $$(n-1)!$$ are actually almost the same, in that $$(n-1)! \times n = n!$$, i.e. they differ only by a factor of $$n$$. The same fact can be applied to see the slightly less obvious fact that $$(n-k)!$$ is just $$(n-k-1)! \times (n-k)$$.