Question & Answer

The value of ${}^{\text{n}}{{\text{C}}_{\text{n}}}$ can be also written as:
A. n
B. 0
C. 1
D. n!

ANSWER Verified Verified
Hint: In order to get the right answer we need to expand ${}^{\text{n}}{{\text{C}}_{\text{n}}}$. As we know ${}^{\text{a}}{{\text{C}}_{\text{b}}}{\text{ = }}\dfrac{{{\text{a!}}}}{{{\text{b!(a - b)!}}}}$. Using the same formula in the given term you will get the right answer.

Complete step-by-step answer:
The given term is ${}^{\text{n}}{{\text{C}}_{\text{n}}}$.
We know that ${}^{\text{a}}{{\text{C}}_{\text{b}}}{\text{ = }}\dfrac{{{\text{a!}}}}{{{\text{b!(a - b)!}}}}$.
So, we can say that:
$ \Rightarrow {}^{\text{n}}{{\text{C}}_{\text{n}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{n!(n - n)!}}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{n!(0)!}}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{n!}}}}{\text{ = 1}}$ (As 0! = 1)
Hence, we come to know that ${}^{\text{n}}{{\text{C}}_{\text{n}}}$= 1.
So, the correct option is C.

Note: To solve this problem we just have to us the formula ${}^{\text{a}}{{\text{C}}_{\text{b}}}{\text{ = }}\dfrac{{{\text{a!}}}}{{{\text{b!(a - b)!}}}}$ knowing that 0! is 1 always and it is always applicable that if the same number terms are present on the above of C and below of C then its value is always 1. Knowing these things you will always get the right answer.