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

Question

Please help me to evaluate combinatorially the following sum: $$\sum_{k=0}^n \binom{n}{k}$$

Thank you.

Answer 1

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.

Answer 2

I will provide an intuition why $$\sum_{k=1}^n \binom{n}{k}=2^n.$$

• firstly, suppose you have $n$ persons, each one has two options either succeed or fail, so you have $\ 2^n $ different combinations

• secondly, let's do it in another way : we will count all the possible combination according to this rule (we choose the ones to succeed, and the rest -who fails- is determined). so we have $\binom{n}{0}$ and the rest fail + $\binom{n}{1}$ and the rest fails. and so on until $\binom{n}{n}$

Answer 3

There are $2^n$ binary words on length $n$, and $\binom{n}{k}$ have $k$ bits set.

By double counting, the sum of the second is the same as the first.

Answer 4

We assume the equality holds and for the last induction step we need to show $$ \sum_{k=0}^{n+1} {n+1 \choose k} = 2^{n+1} $$ Rewrite one of the sums to ${n \choose k-1}$, take out the last and first term of the sums respectively, use that ${n \choose k-1} + {n \choose k} = {n + 1 \choose k}$, the last and first term can be included in the sum again. Similar to the induction proof of the binomial theorem. $$2 \times 2^n = 2 \sum_{k=0}^{n} {n \choose k} = \sum_{k=1}^{n+1} {n \choose k-1} + \sum_{k=1}^{n} {n \choose k} + 1 = 2 + \sum_{k=1}^{n} {n \choose k-1} + {n \choose k} = 2 + \sum_{k=1}^{n} {n + 1 \choose k} = \sum_{k=0}^{n+1} {n + 1 \choose k} $$

The intuition is that when we add one more spice to our repetoire, the number of possible combinations doubles, we can either include it in one of the previous combinations or not.

In the second to last step we add the number of possibilities when it is in the set with the number of possibilities it is not in the set, plus the empty and full set.