Product of Two Binomial Coefficients

An expression with two terms is called the binomial expansion. In the case of higher degree expression, it is difficult to calculate it manually. In these cases, Binomial theorem can be used to calculate it. Binomial theorem is used for the expansion of a binomial expression with a higher degree. Binomial coefficients are the coefficients of the terms in the Binomial expansion. Binomial theorem is proved using the concept of mathematical induction. Apart from Mathematics, Binomial theorem is also used in statistical and financial data analysis.

This Story also Contains

1. Binomial Theorem
2. Binomial Coefficient
3. Product of Two Binomial Coefficients
4. Solved Examples Based on Product of Two Binomial Coefficients

This article is about the product of two Binomial coefficients which falls under the broader category of Binomial Theorem and its applications. It is one of the important topics for competitive exams.

Binomial Theorem

Statement:

If $n$ is any positive integer, then

$ (a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{n - 1} b + \binom{n}{2} a^{n - 2} b^2 + \dots + \binom{n}{n - 1} a b^{n - 1} + \binom{n}{n} b^n $

Proof:

The proof is obtained by applying the principle of mathematical induction.

Let the given statement be:

$ P(n): (a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{n-1} b + \binom{n}{2} a^{n-2} b^2 + \dots + \binom{n}{n-1} a b^{n-1} + \binom{n}{n} b^n $

For $ n = 1 $, we have:

$ P(1): (a + b)^1 = \binom{1}{0} a^1 + \binom{1}{1} b^1 = a + b $

Thus, $ P(1) $ is true.

Suppose $ P(k) $ is true for some positive integer $ k $, i.e.,

$ (a + b)^k = \binom{k}{0} a^k + \binom{k}{1} a^{k-1} b + \binom{k}{2} a^{k-2} b^2 + \dots + \binom{k}{k} b^k$

We shall prove that $ P(k + 1) $ is also true, i.e.,

$ (a + b)^{k + 1} = \binom{k+1}{0} a^{k+1} + \binom{k+1}{1} a^k b + \binom{k+1}{2} a^{k-1} b^2 + \dots + \binom{k+1}{k+1} b^{k+1} $

Now,

$ (a + b)^{k + 1} = (a + b)(a + b)^k $

$ = (a + b) \left[\binom{k}{0} a^k + \binom{k}{1} a^{k-1} b + \binom{k}{2} a^{k-2} b^2 + \dots + \binom{k}{k-1} a b^{k-1} + \binom{k}{k} b^k\right] $

[from (1)]

$ = \binom{k}{0} a^{k+1} + \binom{k}{1} a^k b + \binom{k}{2} a^{k-1} b^2 + \dots + \binom{k}{k-1} a^2 b^{k-1} + \binom{k}{k} a b^k $

$ + \binom{k}{0} a^k b + \binom{k}{1} a^{k-1} b^2 + \binom{k}{2} a^{k-2} b^3 + \dots + \binom{k}{k-1} a b^k + \binom{k}{k} b^{k+1} $

[by actual multiplication]

$ = \binom{k}{0} a^{k+1} + (\binom{k}{1} + \binom{k}{0}) a^k b + (\binom{k}{2} + \binom{k}{1}) a^{k-1} b^2 + \dots + (\binom{k}{k} + \binom{k}{k-1}) a b^k + \binom{k}{k} b^{k+1} $

[grouping like terms]

$ = \binom{k+1}{0} a^{k+1} + \binom{k+1}{1} a^k b + \binom{k+1}{2} a^{k-1} b^2 + \dots + \binom{k+1}{k} a b^k + \binom{k+1}{k+1} b^{k+1}$

(by using $ \binom{k+1}{0} = 1 $, $ \binom{k}{r} + \binom{k}{r-1} = \binom{k+1}{r} $, and $ \binom{k}{k} = 1 = \binom{k+1}{k+1} $)

Thus, it has been proved that $ P(k + 1) $ is true whenever $ P(k) $ is true. Therefore, by the principle of mathematical induction, $ P(n) $ is true for every positive integer $ n $.

Binomial Coefficient

The combination $\binom{n}{r}$ or ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$ occuring in the Binomial theorem is called a Binomial coefficient, where $\binom{n}{r}=C(n, r)={ }^n C_r=\frac{n!}{r!(n-r)!}$.

Product of Two Binomial Coefficients

Consider the identities

$ (1+x)^n=C_0+C_1 x+C_2 x^2+C_3 x^3+\ldots+C_n x^n $

$(1+x)^n=C_0+C_1 x+C_2 x^2+C_3 x^3 \ldots+C_n \cdot x^n $

Multiply these identities we get another identities

$ (1+x)^n(1+x)^n=\left(C_0+C_1 x+C_2 x^2+\ldots \ldots+C_n x^n\right) \times\left(C_0+C_1 x+C_2 x^2+\ldots \ldots+C_n x^n\right) $

$(1+x)^{2 n}=\left(C_0+C_1 x+C_2 x^2+\ldots \ldots+C_n x^n\right) \times\left(C_0+C_1 x+C_2 x^2+\ldots \ldots+C_n x^n\right) $

Compare coefficients of $x^n$ on both sides.

In LHS, coeff. of $x^n=$ coeff. of $x^n$ in $(1+x)^{2 n}={ }^{2 n} C_n$

In RHS. terms containing $x^n$ are

$ C_0 C_n x^n+C_1 C_{n-1} x^n+C_2 C_{n-2} x^n+\ldots \ldots \ldots+C_n C_0 x^n $

$\Rightarrow \quad$ Coeff. of $x^n$ on RHS $=C_0 C_n+C_1 C_{n-1}+C_2 C_{n-2}+\ldots \ldots \ldots+C_n C_0$

Equating the coefficients,

$ C_0 C_n+C_1 C_{n-1}+C_2 C_{n-2}+\ldots \ldots \ldots+C_n C_0=C_0^2+C_1^2+C_2^2+\ldots \ldots \ldots C_n^2={ }^{2 n} C_n $

Shortcut formula

To get the value of $C_0 C_{n-r}+C_1 C_{n-r+1}+C_2 C_{n-r+2}+\ldots \ldots \ldots+C_{n-r} C_0$

Observe that upper indices are constant in each term, and sum of lower indices = constant $=(n-r)$

So this expression equals ${ }^{\text {Sum of upper indices }} C_{\text {sum of lower indices }}$

So the given expression equals ${ }^{n+n} C_{n-r}={ }^{2 n} C_{n-r}$

Recommended Video Based on Product of Two Binomial Coefficients:

Solved Examples Based on Product of Two Binomial Coefficients

Example 1: The value of $r$ for which

${ }^{20} C_r^{20} C_0+{ }^{20} C_{r-1}{ }^{20} C_1+{ }^{20} C_{r-2}{ }^{20} C_2+\ldots \ldots \ldots+{ }^{20} C_0{ }^{20} C_r$ is maximum, is:

1) $15 $

2) $11 $

3) $20$

4) $10 $

Solution:

${ }^{20} C_r^{20} C_0+{ }^{20} C_{r-1}{ }^{20} C_1+{ }^{20} C_{r-2}{ }^{20} C_2+\ldots \ldots \ldots+{ }^{20} C_0{ }^{20} C_r$

Using the shortcut formula

$ ={ }^{20+20} C_r $

$={ }^{40} C_r $

This is maximum when $\mathrm{r}=20$

Example 2: The sum of the series

$\binom{10}{0}\binom{10}{4}+\binom{10}{1}\binom{10}{5}+\binom{10}{2}\binom{10}{6}+\ldots+\binom{10}{6}\binom{10}{10}$ is

1) $\binom{20}{6}$

2) $\binom{20}{7}$

3) $\binom{20}{8}$

4) $\binom{20}{9}$

Solution:

$ \binom{10}{0}\binom{10}{4}+\binom{10}{1}\binom{10}{5}+\binom{10}{2}\binom{10}{6}+\ldots+\binom{10}{6}\binom{10}{10} $

$=\binom{10}{0}\binom{10}{6}+\binom{10}{1}\binom{10}{5}+\binom{10}{2}\binom{10}{4}+\ldots+\binom{10}{6}\binom{10}{0} $

$\left(U \operatorname{sing}{ }^n C_r={ }^n C_{n-r}\right) $

Now upper indices are constant and sum of lower indices is constant

$ =\binom{10+10}{6} $

$=\binom{20}{6} $

Example 3: $\sum_{\text {If }}^{31}\left({ }^{31} \mathrm{C}_{\mathrm{k}}\right)\left({ }^{31} \mathrm{C}_{\mathrm{k}-1}\right)-\sum_{\mathrm{k}=1}^{30}\left({ }^{30} \mathrm{C}_{\mathrm{k}}\right)\left({ }^{30} \mathrm{C}_{\mathrm{k}-1}\right)=\frac{\alpha(60!)}{(30!)(31!)}$, where $\alpha \in \mathrm{R}$, then the value of $16 \alpha$ is equal to

1) $1411 $

2) $1320 $

3) $1615 $

4) $1855 $

Solution:

$ \sum_{\mathrm{k}=1}^{31}{ }^{31} \mathrm{C}_{\mathrm{k}}{ }^{31} \mathrm{C}_{\mathrm{k}-1}-\sum_{\mathrm{k}=1}^{30}{ }^{30} \mathrm{C}_{\mathrm{k}}{ }^{30} \mathrm{C}_{\mathrm{k}-1} $

$=\left({ }^{31} \mathrm{C}_1{ }^{31} \mathrm{C}_0+{ }^{31} \mathrm{C}_2{ }^{31} \mathrm{C}_1+\cdots{ }^{31} \mathrm{C}_{31}{ }^{31} \mathrm{C}_{30}\right)-\left({ }^{30} \mathrm{C}_1{ }^{30} \mathrm{C}_0+{ }^{30} \mathrm{C}_2{ }^{30} \mathrm{C}_1+\cdots{ }^{30} \mathrm{C}_{30}{ }^{30} \mathrm{C}_{20}\right) $

$=\left({ }^{31} \mathrm{C}_1{ }^{31} \mathrm{C}_{31}+{ }^{31} \mathrm{C}_2{ }^{31} \mathrm{C}_{30}+\cdots{ }^{31} \mathrm{C}_{31}{ }^{31} \mathrm{C}_1\right)-\left({ }^{30} \mathrm{C}_1{ }^{30} \mathrm{C}_{30}+{ }^{30} \mathrm{C}_2{ }^{30} \mathrm{C}_{29}+\cdots{ }^{30} \mathrm{C}_{30}{ }^{30} \mathrm{C}_1\right) $

$={ }^{62} \mathrm{C}_{32}-{ }^{60} \mathrm{C}_{31} $

$\left.=\frac{62!}{32!30!}-\frac{60!}{31!29!}=\frac{60![62 \times 61-32 \times 30]}{32!30!}\right) $

$=\frac{60!}{30!31!}\left(\frac{62 \times 61-32 \times 30}{32}\right) $

$ \Rightarrow \alpha=\frac{31 \times 61-16 \times 30}{16} $

$\Rightarrow 16 \alpha=1891-480=1411 $

Hence, the answer is option (1).