Question

The value of $^{15}{C_3}{ + ^{15}}{C_{13}}$ is?
(A) $^{16}{C_3}$
(B) $^{30}{C_{16}}$
(C) $^{15}{C_{10}}$
(D) $^{15}{C_{15}}$

Answer 1

We first discuss the general form of combination and its general meaning with the help of variables. We express the mathematical notion with respect to the factorial form of $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. Then, we place the values of n and r and find the values of combinations separately. We complete the multiplication and add both the values to get to the required answer.

Complete answer:
The general form of combination is $^n{C_r}$ . It’s used to express the notion of choosing r objects out of n objects. The value of $^n{C_r}$ expresses the number of ways the combination of those objects can be done.
The simplified form of the mathematical expression $^n{C_r}$ is $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ .
Here the term $n!$ defines the notion of multiplication of first n natural numbers.
This means $n! = 1 \times 2 \times 3 \times ....n$.
Now, we have to find the value of $^{15}{C_3}{ + ^{15}}{C_{13}}$.
So, we first find the value of $^{15}{C_3}$. We put the values of $n = 15$ and $r = 3$ to get $^{15}{C_3} = \dfrac{{15!}}{{3!\left( {15 - 3} \right)!}}$.
Similarly, we find the value of $^{15}{C_{13}}$. We put the values of $n = 15$ and $r = 13$ to get $^{15}{C_{13}} = \dfrac{{15!}}{{13!\left( {15 - 13} \right)!}}$.
So, we have, $^{15}{C_3}{ + ^{15}}{C_{13}}$
We now solve the factorial values.
$\Rightarrow \dfrac{{15!}}{{3!\left( {15 - 3} \right)!}} + \dfrac{{15!}}{{13!\left( {15 - 13} \right)!}}$
$\Rightarrow \dfrac{{15!}}{{3! \times 12!}} + \dfrac{{15!}}{{13! \times 2!}}$
Taking $15!$ common from both the terms, we get,
$\Rightarrow 15!\left[ {\dfrac{1}{{3! \times 12!}} + \dfrac{1}{{13! \times 2!}}} \right]$
Now, expressing $13! = 13 \times 12!$ and $3! = 3 \times 2!$, we get,
$\Rightarrow 15!\left[ {\dfrac{1}{{3 \times 2! \times 12!}} + \dfrac{1}{{13 \times 12! \times 2!}}} \right]$
Now, taking $12! \times 2!$ common from denominator, we get,
$\Rightarrow \dfrac{{15!}}{{2! \times 12!}}\left[ {\dfrac{1}{3} + \dfrac{1}{{13}}} \right]$
Taking LCM inside the bracket, we get,
$\Rightarrow \dfrac{{15!}}{{2! \times 12!}}\left[ {\dfrac{{13 + 3}}{{39}}} \right]$
Simplifying the calculations, we get,
$\Rightarrow \dfrac{{16 \times 15!}}{{39 \times 2! \times 12!}}$
Now, we know that $39 = 13 \times 3$ and $15! \times 16 = 16!$. So, we get,
$\Rightarrow \dfrac{{16!}}{{\left( {3 \times 2!} \right) \times \left( {13 \times 12!} \right)}}$
Now, we know that $12! \times 13 = 13!$ and $2! \times 3 = 3!$. Hence, we get,
$\Rightarrow \dfrac{{16!}}{{3! \times 13!}}$
Now, we also know that $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. So, if we put $n = 16$ and $r = 3$, we get,
$^{16}{C_3} = \dfrac{{16!}}{{3!\left( {16 - 3} \right)!}} = \dfrac{{16!}}{{3! \times 13!}}$.
Hence, we get,
${ \Rightarrow ^{16}}{C_3}$
Therefore, the value of the expression $^{15}{C_3}{ + ^{15}}{C_{13}}$ is $^{16}{C_3}$.

Therefore, option (A) is the correct answer.

Note:
There are some constraints in the form of $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. Also, we need to remember the fact that the notion of choosing r objects out of n objects is exactly equal to the notion of choosing objects out of n objects. The mathematical expression is $^n{C_{\left( {n - r} \right)}} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}{ = ^n}{C_r}$. The arrangement of those chosen objects is not considered in case of combination. That part is involved in permutation. We must know the simplification and algebraic rules so as to simplify and find the value of expression.