Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to prove $3^n = \sum_{0 \leq j \leq i \leq n} $ $ n \choose i$ $ i \choose j$ using $3^n = \sum_{0 \leq i \leq n} 2^i$ $n \choose i$

share|improve this question
$2 = 1+1$. :-) Also, shouldn't it be $j \leq i$? –  WimC Nov 19 '12 at 16:51
@WimC yes typo, thanks –  xiamx Nov 19 '12 at 16:52

2 Answers 2

up vote 11 down vote accepted

$$ \sum_{0 \leq j \leq i \leq n} {n \choose i} {i \choose j}=\sum_{0 \le i \le n} {n \choose i} \sum_{0 \le j \le i}{i \choose j}=\sum_{0 \le i \le n} {n \choose i} 2^i=3^n. $$

share|improve this answer

Count cardinality of $S = \{(A,B):B \subseteq A \subseteq \left\{1,2,\dots,n\right\}\}$ in two different ways:

Way 1. Each element of $\{1,2,\dots,n\}$ can either be in $A$ and $B$, only in $A$, or in none of $A$ and $B$, so $|S|=3^n$.

Way 2. If $|A|=i$ and $|B|=j$, then there are $n \choose i$ options for $A$ and $i \choose j$ options for $B$, therefore $|S|=\sum_{j \leq i} {n \choose i} {i \choose j}$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.