<s>[INST] 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!
 [/INST] $$2^5$$ is just a way to count the elements in a 5 member set. (For each member of the set you make a binary decision to include it or not)
Now let's say you have a six element set.
Let's look at $$ \binom{6}{1} $$
Fix any member of the 6 element set. Now you have 5 members left. So that is $$ \binom{5}{1} $$ But you could have choosen that fixed member in 6 ways. So that is effectively $$ 6\cdot\binom{5}{1} $$
Repeat the same argument for each binomial and we get:
$$ 6(\binom{5}{1} + \binom{5}{2} +\binom{5}{3} + \binom{5}{4} + \binom{5}{5} + \binom{5}{6}) $$
But what is inside the parenthesis is simply another way  of counting the sub-sets of a set of 5 elements.
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