Question

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

Answer 1

Hint: Firstly, find the values of ${}^{2n}{C_3}$ and ${}^n{P_2}$ by using the formulae ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ and ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ respectively.
Then, substitute the values of ${}^{2n}{C_3}$ and ${}^n{P_2}$ in ${}^{2n}{C_3}:{}^n{P_2} = 10:3$ to form an equation in the terms of n.
Finally, solve the equation to get the value of n.

Complete step-by-step answer:
It is given that ${}^{2n}{C_3}:{}^n{P_2} = 10:3\left( {n \in N} \right)$ .
Now, using ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ , find the value of ${}^{2n}{C_3}$ .
 $
  {}^{2n}{C_3} = \dfrac{{2n!}}{{3!\left( {2n - 3} \right)!}} \\
   = \dfrac{{\left( {2n} \right)\left( {2n - 1} \right)\left( {2n - 2} \right)\left( {2n - 3} \right)!}}{{3!\left( {2n - 3} \right)!}} \\
   = \dfrac{{\left( {2n} \right)\left( {2n - 1} \right)\left( {2n - 2} \right)}}{6} \\
 $
Also, using ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ , find the value of ${}^n{P_2}$ .
 $
  {}^n{P_2} = \dfrac{{n!}}{{\left( {n - 2} \right)!}} \\
   = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)!}}{{\left( {n - 2} \right)!}} \\
   = n\left( {n - 1} \right) \\
 $
Now, to get taking ratio ${}^{2n}{C_3}:{}^n{P_2} = 10:3$
 $\Rightarrow \dfrac{{{}^{2n}{C_3}}}{{{}^n{P_2}}} = \dfrac{{10}}{3}$
Substituting the values of ${}^{2n}{C_3}$ and ${}^n{P_2}$ in above ratio, we get
 $
  \Rightarrow \dfrac{{\dfrac{{2n\left( {2n - 1} \right)\left( {2n - 2} \right)}}{6}}}{{n\left( {n - 1} \right)}} = \dfrac{{10}}{3} \\
  \Rightarrow \dfrac{{ \times 2n\left( {2n - 1} \right)\left( {n - 1} \right)}}{{6n\left( {n - 1} \right)}} = \dfrac{{10}}{3} \\
  \Rightarrow \dfrac{{2\left( {2n - 1} \right)}}{3} = \dfrac{{10}}{3} \\
  \Rightarrow 2n - 1 = \dfrac{{10}}{2} \\
  \Rightarrow 2n - 1 = 5 \\
  \Rightarrow 2n = 6 \\
  \Rightarrow n = \dfrac{6}{2} \\
  \Rightarrow n = 3 \\
 $
Thus, we get \[n = 3\] .
So, option (C) is correct.

Note: Here, reducing the terms \[2n!\] up to \[\left( {2n-3} \right)!\] in ${}^{2n}{C_3}$ and \[n!\] up to \[\left( {n-2} \right)!\] in ${}^n{P_2}$ , will help to solve the question easily.
To reduce any n! up to \[\left( {n-r} \right)!\] , we have to multiply the terms between n! and \[\left( {n-r} \right)!\] in descending order i.e. $n\left( {n - 1} \right)\left( {n - 2} \right)....\left( {n - r - 1} \right)\left( {n - r} \right)!$ .