# If \[{}^n{C_3} = {}^n{C_2}\], then \[n\] is equal to

A) \[2\]

B) \[3\]

C) \[5\]

D) None of these.

Last updated date: 19th Mar 2023

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Answer

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**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.

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