Question

How do you use the Binomial theorem to calculate ${}^{10}{C_7}$?

Answer 1

Binomial Theorem is used to expand the expression which has been raised to any finite power. Using the formula of Binomial theorem in combinations to calculate $^{10}{C_7}$ then try to simplify the values in the formula to easily solve the question.

Complete Step by Step Solution:
The method of expansion of an expression that has been raised to any finite power is called the Binomial theorem. With the help of the Binomial theorem, we can easily calculate the Binomial expression which has been raised to a very large power. It is the powerful tool of expansion that has uses in the Algebra, probability etc.
The binomial expression is the algebraic expression that contains two different terms.
Let $n \in N,x,y, \in R$ then –
${\left( {x + y} \right)^n} = {}^n{\sum {} _{r = 0}}^n{C_r}{x^{n - r}}.{y^r}$ where, -
$^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}} \cdots \left( 1 \right)$
where, $n!$ can be expanded as –
$\to n! = n\left( {n - 1} \right)\left( {n - 2} \right)........ \times 1$
According to the question, we have to calculate the value of $^{10}{C_7}$ , comparing $^{10}{C_7}$ with $^n{C_r}$ , we get –
$n = 10$ and $r = 7$
Putting the above values of $n$ and $r$ in the equation (1), we get –
${ \Rightarrow ^{10}}{C_7} = \dfrac{{10!}}{{\left( {10 - 7} \right)!7!}} \\\ { \Rightarrow ^{10}}{C_7} = \dfrac{{10!}}{{3!7!}} \cdots \left( 2 \right) \\\$
Now, we know that, after expanding $10!$ , we get $10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ , we can also write it as –
$\Rightarrow 10 \times 9 \times 8 \times 7!$ as $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
Putting this value of $10!$ in the equation (2), we get –
${ \Rightarrow ^{10}}{C_7} = \dfrac{{10 \times 9 \times 8 \times 7!}}{{3!7!}}$
Now, canceling the $7!$ as they are present in numerator and denominator both, we get –
${ \Rightarrow ^{10}}{C_7} = \dfrac{{10 \times 9 \times 8}}{{3!}}$
Expand $3!$ present in the denominator to solve further, we get –
${ \Rightarrow ^{10}}{C_7} = \dfrac{{10 \times 9 \times 8}}{{3 \times 2 \times 1}}$
Now, we know that if we divide 9 by 3, we get 3 as quotient and if we divide 8 by 2, we will get quotient as 4, therefore, dividing 9 by 3 and 8 by 2 in the above, we get –
${ \Rightarrow ^{10}}{C_7} = 10 \times 3 \times 4$
Now, multiplying the above solution –
${ \Rightarrow ^{10}}{C_7} = 120$

Hence, the value of $^{10}{C_7}$ using the binomial theorem is $120$.

Note:
Many students can take a long path in this step, $^{10}{C_7} = \dfrac{{10!}}{{3!7!}}$ by not simplifying and not cancelling $7!$ on numerator and denominator. If they solve $10!$ then $3!$ and $7!$, it will make their answer long and take a lot of time. It can also make their answer to be wrong. Therefore, simplify your solution and then try to solve it which will save them much time.