Question

If ${}^{2n}{C_3}:{}^n{P_2} = 10:3$ where $n \in N$ then $n$ is
A. $4$
B. $7$
C. $3$
D. $6$

Answer 1

In this problem, it is given that the ratio of ${}^{2n}{C_3}$ to ${}^n{P_2}$ is equal to the ratio of $10$ to $3$. That is, $\dfrac{{{}^{2n}{C_3}}}{{{}^n{P_2}}} = \dfrac{{10}}{3}$. We need to find the value of $n$. For this, we will use the formula of combination and permutation. We know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ and ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$. We will use this formula in the given relation and we will simplify the expression to get the required value of $n$.

Complete step by step solution: In this problem, it is given that ${}^{2n}{C_3}:{}^n{P_2} = 10:3$. That is, the ratio of ${}^{2n}{C_3}$ to ${}^n{P_2}$ is equal to the ratio of $10$ to $3$. Therefore, we can write this expression as $\dfrac{{{}^{2n}{C_3}}}{{{}^n{P_2}}} = \dfrac{{10}}{3}\; \Rightarrow 3\left[ {{}^{2n}{C_3}} \right] = 10\left[ {{}^n{P_2}} \right] \cdots \cdots \left( 1 \right)$.
Now we are going to use the formula of combination and permutation in equation $\left( 1 \right)$. That is, we are going to use the formula ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ and ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ in equation $\left( 1 \right)$. Therefore,$3\left[ {{}^{2n}{C_3}} \right] = 10\left[ {{}^n{P_2}} \right]$
$\Rightarrow 3\left[ {\dfrac{{\left( {2n} \right)!}}{{3!\left( {2n - 3} \right)!}}} \right] = 10\left[ {\dfrac{{n!}}{{\left( {n - 2} \right)!}}} \right]$
Now we will write $\left( {2n} \right)!$ as $\left( {2n} \right)! = 2n \times \left( {2n - 1} \right) \times \left( {2n - 2} \right) \times \left( {2n - 3} \right)!$. Also we can write $n!$ as $n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right)!$. Therefore, we get
$3\left[ {\dfrac{{2n \times \left( {2n - 1} \right) \times \left( {2n - 2} \right) \times \left( {2n - 3} \right)!}}{{6\left( {2n - 3} \right)!}}} \right] = 10\left[ {\dfrac{{n \times \left( {n - 1} \right) \times \left( {n - 2} \right)!}}{{\left( {n - 2} \right)!}}} \right]$
On cancelling $\left( {2n - 3} \right)!$ and $\left( {n - 2} \right)!$ from above expression and after simplification, we get
$n \times \left( {2n - 1} \right) \times \left( {2n - 2} \right) = 10n \times \left( {n - 1} \right)$
$\Rightarrow n \times \left( {2n - 1} \right) \times 2\left( {n - 1} \right) = 10n \times \left( {n - 1} \right)$
On cancelling $n \times \left( {n - 1} \right)$ from both sides in above expression, we get
$\left( {2n - 1} \right) \times 2 = 10$
$\Rightarrow 2n - 1 = 5$
$\Rightarrow 2n = 5 + 1$
$\Rightarrow 2n = 6$
$\Rightarrow n = \dfrac{6}{2} = 3$
Note that here $n = 3$ is a natural number. Therefore, we can say that $n = 3 \in N$.
Therefore, if ${}^{2n}{C_3}:{}^n{P_2} = 10:3$ then the value of $n$ is $3$.
Therefore, option C is correct.

Note: $N$ is the set of natural numbers. Number of permutations of $n$ objects taken $r$ objects at a time is denoted by ${}^n{P_r}$. It is also denoted by $P\left( {n,r} \right)$ and it is given by $P\left( {n,r} \right) = \dfrac{{n!}}{{\left( {n - r} \right)!}}$. In permutation, order of elements (objects) is important. Number of combinations of $n$ objects taken $r$ at a time is denoted by ${}^n{C_r}$. It is also denoted by $C\left( {n,r} \right)$ and it is given by $C\left( {n,r} \right) = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. In combination, order of elements (objects) is not important.