Product of Two Binomial Coefficients

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).