Prove this identity by making a combinatorial proof, that mean to make a counting argument using photos and/or committees. Solution THE LHS = 0*nC0 + 1*nC1 + 2*nC2 + ... + n*nCn THE RHS = Notice the n is counting the total number of subsets we have from the LHS Furthermore notice that we are also counting all possible subsets. For example nC1 counts all 1 element subsets from a set of n elements. nC2 counts all 2 element subsets from a set of n elements. It is known that for any set with n element that there are 2^n subsets. Why? Well, the most logical way to think about this is saying for each object in the set we can either include it in the subset or not include it. So 2 options for each of n elements is 2 * 2 * 2 ... * 2 (n times) = 2^n. So why do we have 2*n-1 on the RHS? Well as you can see we aren\'t counting the empty set on the LHS because of the 0 * nC0 = 0 Thus, we have shown that the LHS and the RHS count exactly the same thing. This completes the proof..

1. Prove this identity by making a combinatorial proof, that mean to make a counting argument
using photos and/or committees.
Solution
THE LHS = 0*nC0 + 1*nC1 + 2*nC2 + ... + n*nCn THE RHS = Notice the n is
counting the total number of subsets we have from the LHS Furthermore notice that we are also
counting all possible subsets. For example nC1 counts all 1 element subsets from a set of n
elements. nC2 counts all 2 element subsets from a set of n elements. It is known that for any set
with n element that there are 2^n subsets. Why? Well, the most logical way to think about this is
saying for each object in the set we can either include it in the subset or not include it. So 2
options for each of n elements is 2 * 2 * 2 ... * 2 (n times) = 2^n. So why do we have 2*n-1 on
the RHS? Well as you can see we aren't counting the empty set on the LHS because of the 0 *
nC0 = 0 Thus, we have shown that the LHS and the RHS count exactly the same thing. This
completes the proof.