Story proof for $\sum_{k=0}^n {n \choose k} = 2^n$

Question

I found a solution online that uses the Binomial Theorem. Is it possible to prove this without using that theorem?

Well...as $\binom nk$ is the number of subsets of $S[n]={1,...,n}$ with exactly $k$ elements, the left hand is the total number of subsets of $S[n]$. But that's clearly $2^n$. — lulu, May 04 '16 at 18:42

score 7 · Accepted Answer · answered May 04 '16 at 18:42

7

Consider choosing a subset of a set of length $n$. One way is to determine for every item if it will make it or not, giving $2^n$ ways.

The other, choose the size first.

answered May 04 '16 at 18:42

gt6989b

54,422

score 2 · Answer 2 · answered May 04 '16 at 18:54

2

This may depend on what definitions you're using. If $\dbinom n k$ is defined as the number of size-$k$ subsets of a size-$n$ set, and if you know that $2^n$ is the total number of subsets of a size-$n$ set, then the identity just says for every possible size of subsets, if you add up the number of subsets of that size, you get the total number of subsets.

answered May 04 '16 at 18:54

Michael Hardy

1

This is what is meant by a story proof, kudos. – Learning stats by example Mar 21 '20 at 22:23

score 2 · Answer 3 · answered May 04 '16 at 19:25

Using the generation rule from Pascal's triangle: \begin{align} S_n &= \sum_{k=0}^n \binom{n}{k} \\ &= 2 + \sum_{k=1}^{n-1} \binom{n}{k} \\ &= 2 + \sum_{k=1}^{n-1} \left( \binom{n-1}{k-1} + \binom{n-1}{k} \right) \\ &= 2 + \sum_{k=0}^{n-2} \binom{n-1}{k} + \sum_{k=1}^{n-1} \binom{n-1}{k} \\ &= \sum_{k=0}^{n-1} \binom{n-1}{k} + \sum_{k=0}^{n-1} \binom{n-1}{k} \\ &= 2 S_{n-1} \end{align} Further $S_0 = 1$.

This is a homogeneous linear recurrence relation with constant coefficients: \begin{align} S_0 &= 1 \\ S_n &= 2 S_{n-1} \end{align} It has the characteristic polynomial $$ p(t) = t - 2 $$ with root $t = 2$ and general solution $$ S_n = k \, 2^n $$ The initial condition gives $k = 1$. This results in $$ S_n = 2^n $$

score 1 · Answer 4 · answered May 04 '16 at 18:42

HINT:

In how many ways can you choose $k$ things from $n$ things where $0 \le k \le n$ ?

The L.H.S. will come out to be obvious from this argument while for the R.H.S., we have the argument: For each of the $n$ things, you have $2$ choices; you select it or you don't.

score 0 · Answer 5 · 2016-05-04T19:13:35.180

0

Without words: $$\begin{matrix}&1&+&4&+&6&+&4&+&1&+\\ &&&1&+&4&+&6&+&4&+&1\\ =&1&+&5&+&10&+&10&+&5&+&1\\ \end{matrix}$$

$$16+16=32$$

$$0000\\ 1000,0100,0010,0001\\ 1100,1010,1001,0110,0101,0011\\ 1110,1101,1011,0111\\ 1111$$

$$0000,0001,0010,0011,0100,0101,0110,0111,1000,1001,1010,1011,1100,1101,1110,1111$$

edited May 04 '16 at 19:13

answered May 04 '16 at 19:05

1

I believe a few words would not hurt. – Imago May 04 '16 at 19:17
@Imago: would break the rule of the game. – May 04 '16 at 19:19
Yeah, it would. However I find this proof not really obvious. – Imago May 04 '16 at 19:24
@Imago Which one ? – May 04 '16 at 19:24
Yves Daoust: Both proofs :P – Imago May 04 '16 at 19:26
The first is explained by @mvw; the second by @gt6989b and @ SchrodingersCat . – May 04 '16 at 19:28

Story proof for $\sum_{k=0}^n {n \choose k} = 2^n$

5 Answers5