Proof of a common expression in Combinatorics.

Question

I would like to know a mathematical proof of the following expression:

$${\sum\limits_{r=0}^n {n\choose{r}} {=} {2^n}}$$

Thank you!

Try to do it both using the binomial theorem and using a combinatorial argument. — M.B., Feb 06 '14 at 16:37
Shit! That was an easy proof! I must be really dumb! -_- Thank you for the answer! :) — Abir Mukherjee, Feb 06 '14 at 16:49

score 5 · Answer 1 · answered Feb 06 '14 at 16:37

5

There is the Binomial Theorem that says $$ \sum_{k=0}^n\binom{n}{k}x^k=(1+x)^n $$ Plug in $x=1$.

answered Feb 06 '14 at 16:37

robjohn

345,667

Thank you for the solution! – Abir Mukherjee Feb 06 '14 at 16:50

score 4 · Answer 2 · answered Feb 06 '14 at 16:48

Here's a combinatorial proof (just for fun):

The right hand side counts the number of subsets of an $n$-element set. The left hand side does the same, just in a different way. It counts the number of $r$-element subsets for each $r$, and then sums over $n$.

Since they both count the same number of things, just in different ways, they are equivalent.

score 3 · Answer 3 · answered Feb 06 '14 at 16:37

3

Hint:

$$\text{For any pair of real (or complex, if you will) numbers}\;\;a,b\;:$$

$$\;\;\color{red}{(a+b)^n=\sum_{i=0}^n\binom nia^ib^{n-i}}$$

answered Feb 06 '14 at 16:37

DonAntonio

211,718
17
136
287

1

In particular, $a=b=1$ :-) – robjohn Feb 06 '14 at 16:41
2

@robjohn Yeah, well: you may as well write him the complete solution...:) – DonAntonio Feb 06 '14 at 16:45
Thank you for the solution. :) – Abir Mukherjee Feb 06 '14 at 16:51

Proof of a common expression in Combinatorics.

3 Answers3