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Binomial coefficients


We show how Binomial coefficients help simplify expressions. We notice that these coefficients are heavily used in probability calculus. On the other hand, the Binomial theorem serves in computing powers of numbers.

The expression of Binomial Coefficients

For natural numbers $n,k\in\mathbb{N}$ uch that $n\ge k$, we define the Binomial coefficient by \begin{align*} \binom{n}{k}=\frac{n!}{k!(n-k)!}.\end{align*}

The binomial coefficients usually appear in elementary probability.

Exercise: For which integers $p\in{1,2,\cdots,n-1}$, prove that the followining inequality between binomial coefiicients, \begin{align*} \binom{n}{p} < \binom{n}{p+1}.\end{align*}

Proof: To prove this, we first compute \begin{align*}\frac{\binom{n}{p}}{\binom{n}{p+1}}=& \frac{n!}{p!(n-p)!}\times \frac{(p+1)!(n-p-1)!}{n!}\cr &= (p+1)\frac{(n-p-1)!}{(n-p)!}\cr &= \frac{p+1}{n-p}.\end{align*}

We have \begin{align*} \binom{n}{p} < \binom{n}{p+1}&\;\Longleftrightarrow\; \frac{p+1}{n-p} < 1\cr & \;\Longleftrightarrow\; p < \frac{n-1}{2}.\end{align*}

The Binomial theorem

The binomial theorem has many applications in algebra, calculus, and probability. For example, it enters into the definition of the Binomial distribution in probability theory. Here we show a nice application of this Theorem.

Binomial Theorem: For $a$ and $b$ real numbers, and a natural number $n,$ the binomial formula is given by \begin{align*}(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^k b^{n-k}.\end{align*}

The proof of the above sum formula is based on the fact that the multiplication operation in the set of real numbers is commutative, that is $ab=ba$. Otherwise, this formula is not true. For example, we cannot apply the binomial theorem to the sum of two matrices unless these matrices commute with each other.

Exercise: Compute the sum\begin{align*} A_n=\binom{n}{0}+\frac{1}{2}\binom{n}{1}+\cdots+\frac{1}{n+1}\binom{n}{n}.\end{align*} Proof: Using Binome formula, we obtain for any $x\in\mathbb{R}$ and $n\in \mathbb{N},$ \begin{align*}(1+x)^n&= \sum_{k=1}^{n}\binom{n}{k} x^k1^{n-k} \cr &= \binom{n}{0}+\binom{n}{1} x+\cdots+\binom{n}{n} x^n.\end{align*} By taking the integral between $0$ and $1$ in this formula, we obtain \begin{align*} A_n=\frac{2^{n+1}-1}{n+1}.\end{align*}

Exercise: Prove that \begin{align*} &\sum_{k=0}^{n}\binom{n}{k}=2^n,\cr & 4^n\ge \sum_{k=0}^{n} \frac{3^k}{k!}. \end{align*} Proof: For the equality, il suffices to apply the Binomial theorem in the case $a=1$ and $b=1$. Then we obtain \begin{align*} 2^n=(1+1)^n= \sum_{k=0}^{n}\binom{n}{k}  1^k 1^{n-k}=\sum_{k=0}^{n}\binom{n}{k}.\end{align*} On the other hand, to prove the inequality, we use the Binomial theorem in the case of $a=3$ and $b=1$. We have \begin{align*} 4^n=(3+1)^n&=\sum_{k=0}^{n}\binom{n}{k}  3^k 1^{n-k}\cr &=\sum_{k=0}^{n}\binom{n}{k} \frac{3^k}{k!}\frac{n!}{(n-k)!}.\end{align*} Now the result immediately follows from the fact that $n!\ge (n-k)!$ for any $k=1,2,\cdots,n$.

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