Visualize it

$3^N \; = \; (2+1)^N \; = \; \sum_{j=0}^N \binom{N}{j}2^j \; = \; \sum_{j=0}^N \binom{N}{j} \sum_{i=0}^j \binom{j}{i} \; = \; \sum_{0 \le i \le j \le N}\binom{N}{j}\binom{j}{i}$

${10 \choose j}{j \choose i}$ is the number of ways to pick $j$ elements in 10, and then $i$ elements in those $j$ elements. It's the same as the number of ways to color 10 white balls so that $i$ are red and $j-i$ are blue ( $j$ being the number of red and blue balls).

As $i$ and $j$ span all possibilities for some number of red balls and some number of blue balls, the sum is the same as counting all possibilities to paint 10 white balls with red and blue paint: each ball has 3 choices and so the total number is $3^{10}$ .