Question

How do you use the binomial theorem to calculate ${}^8{C_5}$ ?

Answer 1

Hint : To find the value of the ${}^8{C_5}$ term we use the formula binomial expansion and the formula is given as ${(a + b)^n} = {}^n{C_0}{a^n}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + ... + {}^n{C_n}{a^0}{b^n}$ . We apply the binomial expansion where n is 8 and we have to find the ${}^8{C_5}$ . Hence we obtain the solution.

Complete step by step solution:
To solve this question, we use the formula of binomial expansion and after that we use a factorial formula to solve further. Let us consider ${(x + 1)^8}$ to be the term to which we are applying the formula of binomial expansion.
Now we apply binomial expansion to ${\left( {x + 1} \right)^8}$
Here we have n = 8 $a = x$ and $b = 1$ . Substituting all the values in the formula ${(a + b)^n} = {}^n{C_0}{a^n}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + ... + {}^n{C_n}{a^0}{b^n}$
So we have

$${\left( {x + 1} \right)^8} = {}^8{C_0}{\left( x \right)^8}{\left( 1 \right)^0} + {}^8{C_1}{\left( x \right)^{8 - 1}}{\left( 1 \right)^1} + {}^8{C_2}{\left( x \right)^{8 - 2}}{\left( 1 \right)^2} \\\ \+ {}^8{C_3}{\left( x \right)^{8 - 3}}{\left( 1 \right)^3} + {}^8{C_4}{\left( x \right)^{8 - 4}}{\left( 1 \right)^4} + {}^8{C_5}{\left( x \right)^{8 - 5}}{\left( 1 \right)^5} \\\ \+ {}^8{C_6}{\left( x \right)^{8 - 6}}{\left( 1 \right)^6} + {}^8{C_7}{\left( x \right)^{8 - 7}}{\left( 1 \right)^7} + {}^8{C_8}{\left( x \right)^{8 - 8}}{\left( 1 \right)^8} \;$$

We know the formula ${}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$ and we use this formula to simplify the terms and so we have

$$\Rightarrow {\left( {x + 1} \right)^8} = {x^8} + \dfrac{{8!}}{{7!}}{\left( x \right)^7} + \dfrac{{8!}}{{6!2!}}{\left( x \right)^6} + \dfrac{{8!}}{{5!3!}}{\left( x \right)^5} \\\ \+ \dfrac{{8!}}{{4!4!}}{\left( x \right)^4} + \dfrac{{8!}}{{5!3!}}{\left( x \right)^3} + \dfrac{{8!}}{{6!2!}}{\left( x \right)^2} + \dfrac{{8!}}{{7!1!}}\left( x \right) + 1 \;$$

Here we have to find the value of ${}^8{C_5}$
So in the above expansion it is given as
$\Rightarrow {}^8{C_5} = \dfrac{{8!}}{{5!3!}}$
By simplifying we have
For the simplification we need n factorial formula since we factorial therefore the formula is $n! = n \times (n - 1) \times (n - 2) \times ... \times 2 \times 1$ by using this formula we calculate the factorial terms and we have
$\Rightarrow {}^8{C_5} = \dfrac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{(5 \times 4 \times 3 \times 2 \times 1)(3 \times 2 \times 1)}}$
Now divide the similar terms in the both numerator and denominator. So divide $5 \times 4 \times 3 \times 2 \times 1$ both in denominator and denominator we get
$\Rightarrow {}^8{C_5} = \dfrac{{8 \times 7 \times 6}}{{3 \times 2 \times 1}}$
On simplifying we get
$\Rightarrow {}^8{C_5} = 8 \times 7$
On multiplying 8 and 7 we get
$\Rightarrow {}^8{C_5} = 56$
Hence, we have found the value of ${}^8{C_5}$ by using the binomial theorem.
So, the correct answer is “56”.

Note : Here in this type of question we have considered one example of a binomial equation and determined the value of ${}^8{C_5}$ . We can also determine the value by using formula. since they have given us a binomial equation. so we have considered the example. The C represents the combination and it is formulated as ${}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$ .