Alternative ways to write $ k \binom{n}{k} $

Question

I am looking for a way to simplify $ k \binom{n}{k} $. I don't understand what effect the factor $k$ has on the formula. So can anyone please explain what $\ k\binom{n}{k} $ would equate to?

Answer 1

The expression $k\dbinom{n}k$ can be transformed into many equivalent expressions, but one comes up especially often:

$$k\binom{n}k=n\binom{n-1}{k-1}\;.\tag{1}$$

You can verify this computationally, as lab bhattacharjee did in his answer, or combinatorially. For the latter, imagine that you have a group of $n$ athletes, and you want to form a team of $k$ of them and appoint one member of the team to be the captain.

• There are $\binom{n}k$ ways to choose the $k$ members of the team, and once they’ve been chosen, you can select any one of the $k$ to be the captain; these two choices can be made in $k\binom{n}k$ ways altogether.

• Alternatively, you could select any one of the $n$ athletes to be the captain, and then select $k-1$ of the remaining $n-1$ athletes to fill out the rest of the team. These choices can be made in altogether $n\binom{n-1}{k-1}$ ways

Since these are just two different ways of performing the same task, we must have the identity $(1)$.

Answer 2

If integer $k>0,$

$$k\cdot\binom nk=k\cdot\frac{n!}{(n-k)! k!}=k\cdot\frac{ n\cdot(n-1)!}{\{(n-1)-(k-1)\}!\cdot k\cdot (k-1)!}=n\frac{(n-1)!}{\{(n-1)-(k-1)\}!\cdot (k-1)!}$$

I could not understand where to stop