Probability summation

Question

Hello, I do not understand the sums and products section of probability at all. Can you explain how to solve this problem? Any resources where advanced summation, series, product is explained in a simple language?

Answer 1

Use the identity $$ \binom{n}{m}\binom{m}{j}=\binom{n}{j}\binom{n-j}{m-j};\quad n\geq m\geq j\geq 0 $$ to write the general term as $$ \binom{k}{1}\binom{n}{k}\binom{k-1}{2}=n\binom{n-1}{k-1}\binom{k-1}{2}=n\binom{n-1}{2}\binom{n-3}{k-3}\quad k\geq 3 $$ and then apply the binomial theorem.

Answer 2

What are we counting here?

$$\sum_{k=0}^n\binom k1\binom {k-1}2\binom{n}k$$ This counts the ways to select a team of $k$ from $n$ applicants, among which you select a team captain and two lieutenants; summing the count for all possible values of $k$. ($3\leq k\leq n$ actually, because $\binom k1\binom{k-1}2$ is zero for $k$ of $0,1$ or $2$)

So count ways to select a captain and two lieutenants from among the $n$ applicants, multiplying this by the summed count of all the ways to select the rest of the team of $k$ from the $n-3$ remaining. ($0\leq k\leq n-3$)

You can do it! We are selecting $1$ from $n$, $2$ from $n-1$, and for $0\leq k\leq n-3$ selecting $k$ from $n-3$.

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PS: The binomial theorem will be handy simplifying the result. Alternatively the count for ways to select the rest of the team is the count for subsets of a set of size $n-3$.

Answer 3

\begin{align} \sum_{k=0}^n k \binom{k-1}{2}\binom{n}{k} &= \sum_{k=3}^n k \binom{k-1}{2}\binom{n}{k} \\ &= 3\sum_{k=3}^n \frac{k}{3}\binom{k-1}{2}\binom{n}{k}\\ &= 3\sum_{k=3}^n \binom{k}{3}\binom{n}{k}\\ &= 3\sum_{k=3}^n \binom{n}{3}\binom{n-3}{n-k}\\ &= 3 \binom{n}{3}\sum_{k=3}^n\binom{n-3}{n-k}\\ &= 3 \binom{n}{3}\sum_{k=3}^n\binom{n-3}{k-3}\\ &= 3 \binom{n}{3}\sum_{k=0}^{n-3}\binom{n-3}{k}\\ &= 3 \binom{n}{3}2^{n-3}\\ &= 3 \frac{n(n-1)(n-2)}{3!}2^{n-3}\\ &= n(n-1)(n-2)2^{n-4} \end{align}