Product of two Binomial Coefficients - Practice Questions & MCQ

Series Involving Product of two Binomial Coefficients

Example

$
\begin{aligned}
& C_0^2+C_1^2+C_2^2+\ldots C_n^2={ }^{2 n} C_n \quad \text { or } \quad \sum_{r=0}^n C_r^2={ }^{2 n} C_n \\
& \sum_{r=0}^n C_r^2=C_0^2+C_1^2+C_2^2+C_3^2+\ldots C_n^2 \\
& \quad=C_0 \cdot C_0+C_1 \cdot C_1+C_2 \cdot C_2+C_3 \cdot C_3+\ldots+C_n \cdot C_n \\
& =C_0 \cdot C_n+C_1 \cdot C_{n-1}+C_2 \cdot C_{n-2}+C_3 \cdot C_{n-3}+\ldots+C_n \cdot C_0 \\
& \left(U \operatorname{sing}{ }^n C_r={ }^n C_{n-r}\right)
\end{aligned}
$
Here, the sum of lower suffixes in each term is constant
Consider the identities

$
\begin{aligned}
& (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
\end{aligned}
$
Multiply these identities we get another identity.

$
\begin{aligned}
& (1+x)^n(1+x)^n=\left(C_0+C_1 x+C_2 x^2+\ldots \ldots \ldots+C_n x^n\right) \times\left(C_0+C_1 x+C_2 x^2+\ldots \ldots \ldots+C_n x^n\right) \\
& (1+x)^{2 n}=\left(C_0+C_1 x+C_2 x^2+\ldots \ldots \ldots+C_n x^n\right) \times\left(C_0+C_1 x+C_2 x^2+\ldots \ldots \ldots+C_n x^n\right)
\end{aligned}
$
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 the sum of lower indices $=$ constant $=(\mathrm{n}-\mathrm{r})$
So this expression equals Sum of upper indices $C_{\text {sum of lower indices }}$

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

If the difference of lower suffixes of binomial coefficients in each term is the same

Example

Prove that $C_0 C_1+C_1 C_2+C_2 C_3+\ldots \ldots \ldots+C_{n-1} C_n={ }^{2 n} C_{n-1}$
Here the difference between lower indices is constant in each term
We can convert it to a form where the sum of the lower indices in each term is the same, then apply the shortcut formula that we learnt in the previous concept

$
C_0 C_1+C_1 C_2+C_2 C_3+\ldots \ldots \ldots+C_{n-1} C_n
$
Using the relation ${ }^n C_r={ }^n C_{n-r}$ in second coefficients of each term

$
=C_0 C_{n-1}+C_1 C_{n-2}+C_2 C_{n-3}+\ldots \ldots \ldots+C_{n-1} C_0
$
Now observe that the sum of lower indices is constant, so we can apply the shortcut formula that we learnt in the previous concept

$
\begin{aligned}
& ={ }^{n+n} C_{n-1} \\
& ={ }^{2 n} C_{n-1}
\end{aligned}
$