What kind of binomial formula works here?

Question

I need to write the following sum in a simple way (without sigma): $$\sum\limits_{k=2}^{50}\binom{50}{k}\cdot k \cdot (k-1)$$ I tried Newton's binomial theorem and even Vandermonde's identity but I don't see how they fit... I did find a combinatoric question however: "In how many ways can you pick groups from 50 people, such that each group has 2 representatives?"
but I didn't progress far from that... Help?

Answer 1

The reformulation you have found solves it:

$$ \binom{50}{k}\cdot k(k-1) $$ is the number of ways you can select a $k$-person committee out of a pool of 50 persons, and then select a chairman and a vice-chairman from among the committee.

When you sum that over $k=2\ldots 50$ you get the number of ways you can select a committee of any size (but large enough to have a chair and vice) from 50 people and then give it a chairman and a vice-chairman.

Instead of that you can start by selecting a chairman and a vice-chairman among the initial 50, and then select (with complete freedom) which of the remaining 48 people are on the committee too. Thus $$ \sum_{k=2}^{50} \binom{50}{k}\cdot k(k-1) = 50 \cdot 49 \cdot 2^{48} $$

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We can also do that algebraically:

$$ \begin{align} \binom{50}{k}\cdot k(k-1) &= \frac{50!}{k! (50-k)!} k(k-1) \\&= \frac{50!}{(k-2)!(50-k)!} \\&= 50\cdot 49 \cdot \frac{48!}{(k-2)!(48-(k-2))!} \\&= 50 \cdot 49 \cdot \binom{48}{k-2} \end{align} $$ Now pull the common factors $50$ and $49$ out in front of the sum, and apply the formula for the sum of $\binom{48}{j}$s.

Answer 2

Noting that $\binom{50}{k}\cdot k\cdot (k-1) = \frac{50!}{(50-k)!(k-2)!} = 50\cdot 49\cdot \frac{48!}{(48-(k-2))!(k-2)!} = 50\cdot 49\cdot \binom{48}{k-2}$, you get \begin{align*} \sum_{k=2}^{50}\binom{50}{k}\cdot k\cdot (k-1) &= \sum_{k=2}^{50} 50\cdot 49\cdot \binom{48}{k-2}\\ &= 50\cdot 49\cdot \sum_{k=0}^{48}\binom{48}{k}1^k\cdot 1^{48-k}\\ &=50\cdot 49\cdot (1+1)^{48}\\ &= 50\cdot 49\cdot 2^{48}. \end{align*}