Give a combinatorial argument Give a combinatorial argument to show that $$\binom{6}{1} + 2 \binom{6}{2} + 3\binom{6}{3} + 4 \binom{6}{4} + 5 \binom{6}{5} + 6 \binom{6}{6} = 6\cdot2^5$$
Not quite where to starting proving this one. Thanks!

For whatever it's worth, here's a standard probabilistic argument:  The number of ways to get $k$ heads in six trials when tossing a coin is $\dbinom 6 k$.  If all $2^6=64$ sequences are equally likely (as they are when the coin is "fair", i.e. gives heads and tails equally often) then the probability of exactly $k$ successes in $6$ trials is $\dbinom 6 k \cdot \dfrac 1 {64}$.  So the average number of successes in $6$ trials is
$$
\frac{\binom{6}{1} + 2 \binom{6}{2} + 3\binom{6}{3} + 4 \binom{6}{4} + 5 \binom{6}{5} + 6 \binom{6}{6}}{64}.
$$
But the average number of successes in $6$ trials, with probability $1/2$ of success on each trial, is $6\cdot\frac 1 2=3$, so the sum above equals $3$.
On way of showing that the average number of successes with six trials is three is to observe that the average number of successes in one trial is $1/2$, so you're just adding up $1/2$ six times, getting $3$.
