Question

Solve the following using the correct formula:
${}^{{\text{15}}}{{\text{C}}_{\text{3}}} + {}^{{\text{15}}}{{\text{C}}_5} + ... + {}^{{\text{15}}}{{\text{C}}_{15}}$

Answer 1

Hint: In this question we use the theory of permutation and combination. So, before solving this question you need to first recall the basics of this chapter. Here first we use the expansion of \[{\left( {1 + x} \right)^n}\]. And then after we will manipulate the series as asked in the question.

Formula used - ${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n}}!}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$

Complete step-by-step answer:
Given series is-
$\Rightarrow$ ${}^{{\text{15}}}{{\text{C}}_{\text{3}}} + {}^{{\text{15}}}{{\text{C}}_5} + ... + {}^{{\text{15}}}{{\text{C}}_{15}}$
As we know the expansion of \[{\left( {1 + x} \right)^n}\].
$\Rightarrow$\[{\left( {1 + x} \right)^n}\]= ${}^n{{\text{C}}_0} + {}^n{{\text{C}}_1}{x^1} + {}^n{{\text{C}}_2}{x^2} + {}^n{{\text{C}}_{\text{3}}}{x^3} + ... + {}^n{{\text{C}}_n}{x^n}$
$\Rightarrow$\[{\left( {1 + x} \right)^{15}}\]= ${}^{{\text{15}}}{{\text{C}}_0} + {}^{{\text{15}}}{{\text{C}}_1}{x^1} + {}^{{\text{15}}}{{\text{C}}_2}{x^2} + {}^{{\text{15}}}{{\text{C}}_{\text{3}}}{x^3} + ... + {}^{{\text{15}}}{{\text{C}}_{15}}{x^{15}}$ ………….. (1)
Put x=1 in equation (1), we get
$\Rightarrow$\[{\left( {1 + 1} \right)^{15}}\]=${}^{{\text{15}}}{{\text{C}}_0} + {}^{{\text{15}}}{{\text{C}}_1}{\left( 1 \right)^1} + {}^{{\text{15}}}{{\text{C}}_2}{\left( 1 \right)^2} + {}^{{\text{15}}}{{\text{C}}_{\text{3}}}{\left( 1 \right)^3} + ... + {}^{{\text{15}}}{{\text{C}}_{15}}{\left( 1 \right)^{15}}$
$\Rightarrow$${2^{15}}$= ${}^{{\text{15}}}{{\text{C}}_0} + {}^{{\text{15}}}{{\text{C}}_1}{\left( 1 \right)^1} + {}^{{\text{15}}}{{\text{C}}_2}{\left( 1 \right)^2} + {}^{{\text{15}}}{{\text{C}}_{\text{3}}}{x^3} + ... + {}^{{\text{15}}}{{\text{C}}_{15}}{\left( 1 \right)^{15}}$
$\Rightarrow$${2^{15}}$= ${}^{{\text{15}}}{{\text{C}}_0} + {}^{{\text{15}}}{{\text{C}}_1} + {}^{{\text{15}}}{{\text{C}}_2} + {}^{{\text{15}}}{{\text{C}}_{\text{3}}} + ... + {}^{{\text{15}}}{{\text{C}}_{15}}$ …………… (2)
Put x=$ - $1 in equation (1), we get
$\Rightarrow$\[{\left( {1 - 1} \right)^{15}}\]= ${}^{{\text{15}}}{{\text{C}}_0} + {}^{{\text{15}}}{{\text{C}}_1}{\left( { - 1} \right)^1} + {}^{{\text{15}}}{{\text{C}}_2}{\left( { - 1} \right)^2} + {}^{{\text{15}}}{{\text{C}}_{\text{3}}}{\left( { - 1} \right)^3} + ... + {}^{{\text{15}}}{{\text{C}}_{15}}{\left( { - 1} \right)^{15}}$
$\Rightarrow$\[{\left( {1 - 1} \right)^{15}}\]= ${}^{{\text{15}}}{{\text{C}}_0} - {}^{{\text{15}}}{{\text{C}}_1} + {}^{{\text{15}}}{{\text{C}}_2} - {}^{{\text{15}}}{{\text{C}}_{\text{3}}} + ... - {}^{{\text{15}}}{{\text{C}}_{15}}$
$\Rightarrow$0 = ${}^{{\text{15}}}{{\text{C}}_0} - {}^{{\text{15}}}{{\text{C}}_1} + {}^{{\text{15}}}{{\text{C}}_2} - {}^{{\text{15}}}{{\text{C}}_{\text{3}}} + ... - {}^{{\text{15}}}{{\text{C}}_{15}}$ …………… (3)
Now, we subtract equation (3) from equation (2), we will get-
$\Rightarrow$${2^{15}} - 0$= $2\left( {{}^{{\text{15}}}{{\text{C}}_1} + {}^{{\text{15}}}{{\text{C}}_{\text{3}}} + {}^{{\text{15}}}{{\text{C}}_5} + ... + {}^{{\text{15}}}{{\text{C}}_{15}}} \right)$
$\Rightarrow$$\dfrac{{{2^{15}}}}{2}$= $\left( {{}^{{\text{15}}}{{\text{C}}_1} + {}^{{\text{15}}}{{\text{C}}_{\text{3}}} + {}^{{\text{15}}}{{\text{C}}_5} + ... + {}^{{\text{15}}}{{\text{C}}_{15}}} \right)$
$\Rightarrow$${2^{14}}$= $\left( {{}^{{\text{15}}}{{\text{C}}_1} + {}^{{\text{15}}}{{\text{C}}_{\text{3}}} + {}^{{\text{15}}}{{\text{C}}_5} + ... + {}^{{\text{15}}}{{\text{C}}_{15}}} \right)$
As we know, ${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n}}!}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$
$\Rightarrow$${2^{14}}$= $\left( {15 + {}^{{\text{15}}}{{\text{C}}_{\text{3}}} + {}^{{\text{15}}}{{\text{C}}_5} + ... + {}^{{\text{15}}}{{\text{C}}_{15}}} \right)$ $\left[ {{}^{{\text{15}}}{{\text{C}}_{\text{1}}} = \dfrac{{15!}}{{14! \cdot 1!}}} \right]$
$\Rightarrow$${2^{14}} - 15$= $\left( {{}^{{\text{15}}}{{\text{C}}_{\text{3}}} + {}^{{\text{15}}}{{\text{C}}_5} + ... + {}^{{\text{15}}}{{\text{C}}_{15}}} \right)$

Therefore, ${}^{{\text{15}}}{{\text{C}}_{\text{3}}} + {}^{{\text{15}}}{{\text{C}}_5} + ... + {}^{{\text{15}}}{{\text{C}}_{15}}$ will be equal to ${2^{14}} - 15$.

Note:
Here, C stands for combination. Combinations are the way of selecting the objects or numbers from a group of objects or collection, in such a way that the order of the objects does not matter. For example, suppose we have a set of three letters: A, B, and C. Each possible selection would be an example of a combination.