Question

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

Answer 1

Hint: 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 \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!\;\;r!\;}}\] where \[n\] is the total items in the set and \[r\] is the number of items taken for the permutation to find the value of \[n\].

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 \[1\] i.e., \[n! = n \times \left( {n - 1} \right) \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times - - - \times 2 \times 1\] . The factorial of \[n\] is denoted as \[n!\].
Given \[{}^n{C_3} = {}^n{C_2}\]---(1)

Using the combination formula \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!\;\;r!\;}}\]in the equation (1), we get
\[\dfrac{{n!}}{{\left( {n - 3} \right)!\;\;3!\;}} = \dfrac{{n!}}{{\left( {n - 2} \right)!\;\;2!\;}}\]---(2)
Since \[n! = n\left( {n - 1} \right)\left( {n - 2} \right)!\] and \[n! = n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)!\] then the equation (2) becomes
\[\dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)!}}{{\left( {n - 3} \right)!\;\;3!\;}} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)!}}{{\left( {n - 2} \right)!\;\;2!\;}}\]--(3)
Since \[3! = 3 \times 2 \times 1 = 6\] and \[2! = 2\] then the equation (3) becomes
\[\dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)!}}{{\left( {n - 3} \right)!\;6\;\;}} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)!}}{{\left( {n - 2} \right)!\;\;2\;}}\]--(4)
Simplifying the equation (4), we get
\[\dfrac{{\left( {n - 2} \right)}}{{6\;}} = \dfrac{1}{{2\;}}\]
\[ \Rightarrow n - 2 = 3\]
\[ \Rightarrow n = 5\]
Hence, $n=5$. 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 \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] where \[n\] is the total items in the set and \[r\] is the number of items taken for the permutation.