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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...

If 2nC3:nP2=10:3(n∈N){}^{2n}{C_3}:{}^n{P_2} = 10:3\left( {n \in N} \right) , then n is
A.4
B.7
C.3
D.6

Explanation

Solution

Firstly, find the values of 2nC3{}^{2n}{C_3} and nP2{}^n{P_2} by using the formulae nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} and nPr=n!(n−r)!{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}} respectively.
Then, substitute the values of 2nC3{}^{2n}{C_3} and nP2{}^n{P_2} in 2nC3:nP2=10:3{}^{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 2nC3:nP2=10:3(n∈N){}^{2n}{C_3}:{}^n{P_2} = 10:3\left( {n \in N} \right) .
Now, using nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} , find the value of 2nC3{}^{2n}{C_3} .
2nC3=2n!3!(2n−3)! =(2n)(2n−1)(2n−2)(2n−3)!3!(2n−3)! =(2n)(2n−1)(2n−2)6  {}^{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 nPr=n!(n−r)!{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}} , find the value of nP2{}^n{P_2} .
nP2=n!(n−2)! =n(n−1)(n−2)!(n−2)! =n(n−1)  {}^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 2nC3:nP2=10:3{}^{2n}{C_3}:{}^n{P_2} = 10:3
⇒2nC3nP2=103\Rightarrow \dfrac{{{}^{2n}{C_3}}}{{{}^n{P_2}}} = \dfrac{{10}}{3}
Substituting the values of 2nC3{}^{2n}{C_3} and nP2{}^n{P_2} in above ratio, we get
⇒2n(2n−1)(2n−2)6n(n−1)=103 ⇒×2n(2n−1)(n−1)6n(n−1)=103 ⇒2(2n−1)3=103 ⇒2n−1=102 ⇒2n−1=5 ⇒2n=6 ⇒n=62 ⇒n=3  \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=3n = 3 .
So, option (C) is correct.

Note: Here, reducing the terms 2n!2n! up to (2n−3)!\left( {2n-3} \right)! in 2nC3{}^{2n}{C_3} and n!n! up to (n−2)!\left( {n-2} \right)! in nP2{}^n{P_2} , will help to solve the question easily.
To reduce any n! up to (n−r)!\left( {n-r} \right)! , we have to multiply the terms between n! and (n−r)!\left( {n-r} \right)! in descending order i.e. n(n−1)(n−2)....(n−r−1)(n−r)!n\left( {n - 1} \right)\left( {n - 2} \right)....\left( {n - r - 1} \right)\left( {n - r} \right)! .