Why the Sum of Binomial Coefficients Equals 2 to the n

Adds up nC0 + nC1 + … + nCn, the sum of row n of Pascal's triangle. It always equals 2ⁿ, the number of subsets of an n-element set.

Explanation

The binomial coefficient $\binom{n}{k}$ counts the ways of choosing $k$ items out of $n$ . Add these up across every $k$ from 0 to $n$ and the total always lands on $2^n$ exactly. In terms of Pascal's triangle, you are summing row $n$ .

\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = 2^n

$n$ — an integer of 0 or greater
Sum — the binomial coefficients added together
$2^n$ — shown alongside for comparison; the two should agree

The identity is clearest when you count subsets two ways. The left side sorts the subsets of an $n$ -element set by size and tallies each group separately. The right side counts the same subsets by walking through the $n$ elements and deciding, for each one, whether to include it: two choices, $n$ times over. Same collection, two ways of counting it, so the totals must match.

Example

Take $n = 10$ . Row 10 of Pascal's triangle reads

1,\ 10,\ 45,\ 120,\ 210,\ 252,\ 210,\ 120,\ 45,\ 10,\ 1

Adding those eleven numbers gives 1024, which is exactly $2^{10}$ .

Notes

$n$ must be a whole number of 0 or greater. At $n = 0$ the sum is $\binom{0}{0} = 1$ , matching $2^0 = 1$ .
The identity is just the binomial theorem $(1 + x)^n = \sum_k \binom{n}{k} x^k$ evaluated at $x = 1$ .
Every row is symmetric, because $\binom{n}{k} = \binom{n}{n-k}$ . Above, the values mirror around the central 252.
This calculator allows $n$ up to 53. Double-precision arithmetic holds integers exactly only up to $2^{53}$ , and beyond that the sum and $2^n$ would print as two different numbers. For larger $n$ , use the power calculator to get $2^n$ .