Question

How do you evaluate ${}^5{C_2} + {}^5{C_1}$?

Answer 1

Hint: The given question is of Combination. Let us first know what a combination is in context to mathematics. A combination is one of the processes of selection of all or part of a group of things or objects, without regard to the order in which things are being selected. It is expressed in the form of ${}^n{C_r}$. The number of combinations of $n$ objects taken $r$ at a time is expressed as ${}^n{C_r}$, where
${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} = \dfrac{{n(n - 1)(n - 2).....(n - r + 1)}}{{r(r - 1)(r - 2)......(2)(1)}}$
We will use the above given formula of ${}^n{C_r}$ to solve the question.

Complete Step by Step Solution:
The given expression is ${}^5{C_2} + {}^5{C_1}$. The formula which we will use to solve the expression is stated as:
 ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
We will find the values of ${}^5{C_2}$ and ${}^5{C_1}$ separately and then add them up together to get the final answer. Therefore,
For ${}^5{C_2}$:
We have $n = 5$ and $r = 2$. On substituting these values of $n$ and $r$ in the formula, we will get
$ \Rightarrow {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
$ \Rightarrow {}^5{C_2} = \dfrac{{5!}}{{2!(5 - 2)!}}$
On further simplifying, we will get
 $ \Rightarrow {}^5{C_2} = \dfrac{{5!}}{{2!3!}}$
It must be known that the factorial of a number $n$ is expressed as $n! = n(n - 1)(n - 2)(n - 3)....(2)(1)$. So from this formula for $n!$, we will find the values of $5!$, $2!$ and $3!$, such that
$ \Rightarrow {}^5{C_2} = \dfrac{{(5 \times 4 \times 3 \times 2 \times 1)}}{{(2 \times 1)(3 \times 2 \times 1)}}$
$ \Rightarrow {}^5{C_2} = \dfrac{{120}}{{12}}$
On further simplifying the obtained fraction into its simplest form, we will get
$ \Rightarrow {}^5{C_2} = 10$
Hence we have ${}^5{C_2} = 10$.
Now, for ${}^5{C_1}$:
We have $n = 5$ and $r = 1$. On substituting these values of $n$ and $r$ in the formula, we will get
$ \Rightarrow {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
$ \Rightarrow {}^5{C_1} = \dfrac{{5!}}{{2!(5 - 1)!}}$
On further simplifying, we will get
 $ \Rightarrow {}^5{C_1} = \dfrac{{5!}}{{2!4!}}$
Since for a number $n$, $n! = n(n - 1)(n - 2)(n - 3)....(2)(1)$. So from this formula for $n!$, we will find the values of $5!$, $2!$ and $4!$, such that
$ \Rightarrow {}^5{C_1} = \dfrac{{(5 \times 4 \times 3 \times 2 \times 1)}}{{(2 \times 1)(4 \times 3 \times 2 \times 1)}}$
$ \Rightarrow {}^5{C_1} = \dfrac{{120}}{{48}}$
On further simplifying the obtained fraction into its simplest form, we will get
$ \Rightarrow {}^5{C_1} = \dfrac{5}{2}$
Hence we have ${}^5{C_1} = \dfrac{5}{2}$.
Since ${}^5{C_2} = 10$ and ${}^5{C_1} = \dfrac{5}{2}$, we will find the value of ${}^5{C_2} + {}^5{C_1}$ as
$ \Rightarrow {}^5{C_2} + {}^5{C_1} = 10 + \dfrac{5}{2}$
On further simplifying, we will get
$ \Rightarrow {}^5{C_2} + {}^5{C_1} = \dfrac{{25}}{2}$

Hence, on evaluating ${}^5{C_2} + {}^5{C_1}$, we get $\dfrac{{25}}{2}$ as the answer.

Note:
It must be known that the factorial of $0$, i.e. $0!$$ = 1$. Therefore the value of ${}^n{C_0}$ can be stated as:
\[\]$ \Rightarrow {}^n{C_0} = \dfrac{{n!}}{{0!(n - 0)!}}$
$ \Rightarrow {}^n{C_0} = \dfrac{{n!}}{{0!n!}}$
Since the $n!$ in both the numerator and denominator divides each other, we are left with
$ \Rightarrow {}^n{C_0} = \dfrac{1}{{0!}}$
Since $0! = 1$, therefore we have
$ \Rightarrow {}^n{C_0} = 1$