Counting arguments question about sums of binomial coefficients

Question

Use counting arguments to prove these identities: I don't know how to type this: it is two numbers in brackets the first on top of the other, but there is no fraction line. Here is an image of both of the identities.

$${n \choose 0} + {n \choose 1} + {n \choose 2} + \cdots + {n \choose n} = 2^n$$ $$ {n \choose 1} + 2 {n \choose 2} + 3 {n \choose 3} + \cdots + n{n\choose n} = n 2^{n-1}$$

Your image doesn't load. Please post code or equations (not images). — David G. Stork, Jun 04 '16 at 12:56
So the equations are something like this (instead of beside pretend the first is above the second) (n 0) + (n 1) + (n 2) + ... + (n n) = 2^n (n 1) + 2(n 2) + 3(n 3) + ... + n(n n) = n2^(n-1) — Huk Jin, Jun 04 '16 at 13:01
Thank you so very much I just didn't know what those are and what they mean and how to type them — Huk Jin, Jun 04 '16 at 13:08
$\binom{n}{m}$ has the meaning that it is the number of unique groups of $m$ you can choose from $n$ distinct items. If you sum up all possible groups then you get the total sum of the different ways you can choose items from the group of $n$. If you place each item side-by-side, you can mark it with a $0$ or a $1$ meaning it is selected or it isn't. There are $2^n$ possible ways to assign $0$'s and $1$'s and thus $2^n$ possible ways of picking items. The sum of all of the different groups should add to this total. — Jared, Jun 04 '16 at 13:13
You are asking two questions in your post, it would be better to ask each of them separately. However, both of them have already been asked on this site, see Combinatorial proof of $\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}$.. and Evaluate $\sum\limits_{k=0}^n \binom{n}{k}$ combinatorially. — Martin Sleziak, Jul 26 '16 at 10:28

score 0 · Answer 1 · answered Jun 04 '16 at 13:39

(as per my comment): $\binom{m}{n}$ has the meaning that it is the number of unique groups of $m$ you can choose from $n$ distinct items. If you sum up all possible groups then you get the total sum of the different ways you can choose items from the group of $n$. If you place each item side-by-side, you can mark it with a $0$ or a $1$ meaning it is selected or it isn't. There are $2^n$ possible ways to assign $0$'s and $1$'s and thus $2^n$ possible ways of picking items. The sum of all of the different groups should add to this total.

For the second equation, it might help to understand it as a weighted sum (part of an average). The sum:

$$ \sum_0^n k\binom{n}{k} $$

Makes up part of the computation for the expected size of a group if you randomly select items from the total. This expected value is half of the size of the total. Thus:

$$ \frac{\sum_0^n k\binom{n}{k}}{\sum_0^n \binom{n}{k}} = \frac{n}{2} $$

With the result that $\sum_0^n\binom{n}{k} = 2^n$, this gives:

$$ \sum_0^n k\binom{n}{k} = \frac{n}{2}2^n = n2^{n - 1} $$

score 0 · Answer 2 · answered Jun 04 '16 at 13:46

The first equation can be interpreted combinatorically as: The number of subsets of a set with $n$ elements (i.e. $2^n$) is equal to the sum of the numbers of $k$-element subsets of the set (i.e. $\binom{n}{k}$). This statement is somewhat trivial: If you count all subsets of size $k$ and do this for every $k\in\{0,\dots,n\}$, then you are counting all subsets (each exactly once). If you want a more technical proof, apply the binomial theorem which gives us

$$2^n = (1+1)^n = \sum_{k=0}^{n}\binom{n}{k}1^n1^{k-n} = \sum_{k=0}^{n}\binom{n}{k}.$$

For the second equation, we may use the binomial theorem again:

$$n2^{n-1} = n(1+1)^{n-1} = n\sum_{k=0}^{n-1}\binom{n-1}{k}1^{n-1}1^{k-n+1} = \sum_{k=0}^{n-1}n\binom{n-1}{k} = \sum_{k=0}^{n-1}(n-k)\binom{n}{k} = \sum_{k=0}^{n-1}(n-k)\binom{n}{n-k} = \sum_{k=1}^{n}k\binom{n}{k}.$$

Note that we have used $n\binom{n-1}{k} = n\frac{(n-1)!}{k!(n-k-1)!} = \frac{n!}{(n-k-1)!k!} = (n-k)\frac{n!}{(n-k)!k!} = (n-k)\binom{n}{k}$, as well as $\binom{n}{k}=\binom{n}{n-k}$. The last step of the equation follows from turning around the order of summation.

Counting arguments question about sums of binomial coefficients

2 Answers2