Evaluate $\sum\limits_{k=0}^n \binom{n}{k}$ combinatorially

We use the powerful strategy of counting the same thing in two different ways. We have a set $S$ of $n$ spices. We ask how many different subsets this set has.

Line up the spices in order on a shelf. Go gradually down the shelf, saying yes or no to each spice in turn. At each spice, we have two choices. So there is a total of $2^n$ choices, and hence $S$ has $2^n$ subsets.

For any $k$, there are by definition $\binom{n}{k}$ ways to choose $k$ spices from the set $S$. So $S$ has $\binom{n}{k}$ subsets with $k$ elements. Summing over all $k$ from $0$ to $n$ gives us a different way of counting all the subsets.

Both counting methods are correct, so they must give the same answer. It follows that $$\sum_{k=1}^n \binom{n}{k}=2^n.$$

Remark: Bhaskara once asked the following question. There are $6$ basic flavours (sour, sweet, bitter, and so on). How many different-flavoured dishes can one make by using flavours selected from these? He gave the answer $63$, leaving out the empty set of flavours. He did not know about English cooking.