Question

If ${}^{n}{{C}_{15}}={}^{n}{{C}_{8}}$ , then the value of ${}^{n}{{C}_{21}}$ is
(a) 254
(b) 250
(c) 253
(d) none of these

Answer 1

We can say that ${}^{n}{{C}_{15}}={}^{n}{{C}_{8}}$ so, values of ‘n’, ‘r’ are same. We can write it as ${}^{n}{{C}_{n-15}}={}^{n}{{C}_{8}}$ . After this we will directly compare ’r’ and get the value of n. Then putting the value in ${}^{n}{{C}_{21}}$ , and using the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\cdot r!}$ , we will get the answer.

Complete step-by-step answer :
Here, we are given that ${}^{n}{{C}_{15}}={}^{n}{{C}_{8}}$ . So, we can say that it is in form ${}^{n}{{C}_{r}}$ so, both values of r are the same. We can write it as ${}^{n}{{C}_{n-15}}={}^{n}{{C}_{8}}$ .
So, we can directly compare value of r i.e. $n-15=8$
On solving this, we will get the value of n to be $n=15+8=23$ .
Now, we will put this value of n in ${}^{n}{{C}_{21}}$ . We will use the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\cdot r!}$ where n is 23 and r is 21. So, we can write it as
${}^{23}{{C}_{21}}=\dfrac{23!}{\left( 23-21 \right)!\cdot 21!}$
On further solving, we get as
${}^{23}{{C}_{21}}=\dfrac{23!}{\left( 2 \right)!\cdot 21!}$
${}^{23}{{C}_{21}}=\dfrac{23\times 22\times 21!}{2\cdot 1\cdot 21!}$
Now, we will cancel the same term we get as
${}^{23}{{C}_{21}}=\dfrac{23\times 22}{2\cdot 1}$
Thus, we get answer as
${}^{23}{{C}_{21}}=\dfrac{506}{2}=253$
Hence, option (c) is the correct answer.

Note : We can also write it as ${}^{n}{{C}_{15}}={}^{n}{{C}_{n-8}}$ instead of ${}^{n}{{C}_{n-15}}={}^{n}{{C}_{8}}$ , then also we will get n as 23 on solving. Also, if we apply the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\cdot r!}$ to ${}^{n}{{C}_{15}}={}^{n}{{C}_{8}}$ then it will be very tedious and complex to solve and might lead to wrong answer. Thus, it is easy to directly compare ‘r’ values and get the value of n easily.