Question

How do you find the binomial coefficient of $\left( \begin{align}
  & 10 \\
 & 4 \\
\end{align} \right)$ ?

Answer 1

Hint: We write the coefficient of ${{x}^{r}}$ in the expansion of ${{\left( 1+x \right)}^{n}}$ as ${}^{n}{{C}_{r}}$ another notation of ${}^{n}{{C}_{r}}$ is $\left( \begin{align}
  & n \\
 & r \\
\end{align} \right)$ the value of ${}^{n}{{C}_{r}}$ or $\left( \begin{align}
  & n \\
 & r \\
\end{align} \right)$ is equal to $\dfrac{n!}{r!\left( n-r \right)!}$ . Using this expression we can solve the question.

Complete step by step answer:
We have to find the value of binomial coefficient of $\left( \begin{align}
  & 10 \\
 & 4 \\
\end{align} \right)$
$\left( \begin{align}
  & n \\
 & r \\
\end{align} \right)$ Represents the coefficient of ${{x}^{r}}$ in the expansion of ${{\left( 1+x \right)}^{n}}$ , the value of $\left( \begin{align}
  & n \\
 & r \\
\end{align} \right)$ is equal to $\dfrac{n!}{r!\left( n-r \right)!}$ . Another notation of $\left( \begin{align}
  & n \\
 & r \\
\end{align} \right)$ is ${}^{n}{{C}_{r}}$ .
Now we can write the value of $\left( \begin{align}
  & 10 \\
 & 4 \\
\end{align} \right)$ we have replace n with 10 and r with 4 in the formula $\dfrac{n!}{r!\left( n-r \right)!}$
So the value of $\left( \begin{align}
  & 10 \\
 & 4 \\
\end{align} \right)$ = $\dfrac{10!}{4!\left( 10-4 \right)!}$
$\dfrac{10!}{4!\left( 10-4 \right)!}=\dfrac{10!}{4!6!}$
We know that factorial of any positive integer is the product of all integers from 1 to the integer
For example $n!=1\times 2\times ......\times n$
So the value of $10!=1\times 2\times 3\times ......\times 9\times 10$
Similarly we can write $4!=1\times 2\times 3\times 4$ and $6!=1\times 2\times ......\times 6$
So solving the equation we get
$\dfrac{10!}{4!6!}=2100$
So the value of $\left( \begin{align}
  & 10 \\
 & 4 \\
\end{align} \right)$ is equal to 2100.

Note:
We already know that $\left( \begin{align}
  & n \\
 & r \\
\end{align} \right)$ represent the coefficient of ${{x}^{n}}$ in the expansion of ${{\left( 1+x \right)}^{n}}$ .
$\left( \begin{align}
  & n \\
 & r \\
\end{align} \right)$ also represents the total possible combination of r objects out of n different objects .
Let’s understand this with an example, let’s assume there are 30 different people in a class . We have to find the total number of ways we can make a group of 4 people. In this example we have find total possible combination of 4 people out 30 , the answer is $\left( \begin{align}
  & 30 \\
 & 4 \\
\end{align} \right)$ which is equal to
$\dfrac{30!}{4!26!}$ . Keep in mind that in the above example we only evaluate the possible combination not the permutation we did not calculate how many ways the 4 people will arrange themselves , formula for permutation is $\left( \begin{align}
  & n \\
 & r \\
\end{align} \right)r!$ .