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Question: If \[{}^n{C_3} = {}^n{C_2}\], then \[n\] is equal to A) \[2\] B) \[3\] C) \[5\] D) None of t...

If nC3=nC2{}^n{C_3} = {}^n{C_2}, then nn is equal to
A) 22
B) 33
C) 55
D) None of these.

Explanation

Solution

First we have to know a combination is a mathematical technique that determines the number of possible arrangements in a collection of items where we select the items in any order. Using the combination formula nCr=n!(n−r)!    r!  {}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!\;\;r!\;}} where nn is the total items in the set and rr is the number of items taken for the permutation to find the value of nn.

Complete step by step solution:
The factorial of a natural number is a number multiplied by "number minus one", then by "number minus two", and so on till 11 i.e., n!=n×(n−1)×(n−1)×(n−2)×−−−×2×1n! = n \times \left( {n - 1} \right) \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times - - - \times 2 \times 1 . The factorial of nn is denoted as n!n!.
Given nC3=nC2{}^n{C_3} = {}^n{C_2}---(1)

Using the combination formula nCr=n!(n−r)!    r!  {}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!\;\;r!\;}}in the equation (1), we get
n!(n−3)!    3!  =n!(n−2)!    2!  \dfrac{{n!}}{{\left( {n - 3} \right)!\;\;3!\;}} = \dfrac{{n!}}{{\left( {n - 2} \right)!\;\;2!\;}}---(2)
Since n!=n(n−1)(n−2)!n! = n\left( {n - 1} \right)\left( {n - 2} \right)! and n!=n(n−1)(n−2)(n−3)!n! = n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)! then the equation (2) becomes
n(n−1)(n−2)(n−3)!(n−3)!    3!  =n(n−1)(n−2)!(n−2)!    2!  \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)!}}{{\left( {n - 3} \right)!\;\;3!\;}} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)!}}{{\left( {n - 2} \right)!\;\;2!\;}}--(3)
Since 3!=3×2×1=63! = 3 \times 2 \times 1 = 6 and 2!=22! = 2 then the equation (3) becomes
n(n−1)(n−2)(n−3)!(n−3)!  6    =n(n−1)(n−2)!(n−2)!    2  \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)!}}{{\left( {n - 3} \right)!\;6\;\;}} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)!}}{{\left( {n - 2} \right)!\;\;2\;}}--(4)
Simplifying the equation (4), we get
(n−2)6  =12  \dfrac{{\left( {n - 2} \right)}}{{6\;}} = \dfrac{1}{{2\;}}
⇒n−2=3\Rightarrow n - 2 = 3
⇒n=5\Rightarrow n = 5
Hence, n=5n=5. So, Option (C) is correct.

Note:
Note that a permutation is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. The formula for a permutation is nPr=n!(n−r)!{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}} where nn is the total items in the set and rr is the number of items taken for the permutation.