Properties of Binomial Coefficients

This page gathers together some of the simpler and more common identities concerning binomial coefficients.

= Theorems =

Sum of All Coefficients

 * $$\sum_{i=0}^n \binom n i = 2^n$$

Alternating Sum and Difference of All Coefficients

 * $$\sum_{i=0}^n \left({-1}\right)^i \binom n i = 0$$ for all $$n > 0$$.

= Proofs =

Proof by induction
For all $$n \in \N$$, let $$P \left({n}\right)$$ be the proposition $$\sum_{i=0}^n \binom n i = 2^n$$.


 * $$P(0)$$ is true, as this just says $$\binom 0 0 = 1$$. This holds by definition.

Basis for the Induction

 * $$P(1)$$ is true, as this just says $$\binom 1 0 + \binom 1 1 = 2$$. This also holds by definition.

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 1$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So this is our induction hypothesis:

$$\sum_{i=0}^k \binom {k} {i} = 2^k$$.

Then we need to show:

$$\sum_{i=0}^{k+1} \binom {k+1} {i} = 2^{k+1}$$.

Induction Step
This is our induction step:

$$ $$ $$ $$ $$ $$ $$

So $$P \left({k}\right) \Longrightarrow P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore $$\forall n \in \N: \sum_{i=0}^n \binom n i = 2^n$$.

Alternative Proof
Let $$S$$ be a set with $$n$$ elements.

From the definition of $r$-combination, $$\sum_{i=0}^n \binom n i$$ is the total number of subsets of $$S$$.

Hence $$\sum_{i=0}^n \binom n i$$ is equal to the cardinality of the power set of $$S$$.

Hence the result.

Proof of Alternating Sum and Difference of All Coefficients
$$ $$ $$

We note:
 * $$\binom n {0} = \binom {n-1} {0} = 1$$ so $$\binom n {0} - \binom {n-1} {0}$$;
 * $$\left({-1}\right)^{n-1} \binom {n-1} {n-1} = - \left({-1}\right)^n \binom n n = \left({-1}\right)^n$$ so $$\left({-1}\right)^{n-1} \binom {n-1} {n-1} + \left({-1}\right)^n \binom n n = 0$$.

Hence the result.