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# If ${}^n{C_3} = {}^n{C_2}$, then $n$ is equal toA) $2$B) $3$C) $5$D) None of these.

Last updated date: 14th Jul 2024
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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.