Question

If we have given ${}^{n}{{C}_{3}}+{}^{n}{{C}_{4}}>{}^{n+1}{{C}_{3}}$, then which of the following is true?
(a) n > 6
(b) n > 7
(c) n < 6
(d) None of these.

Answer 1

Hint: We start solving the problem by substituting ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ in place of ${}^{n}{{C}_{3}}$, ${}^{n}{{C}_{4}}$ and ${}^{n+1}{{C}_{3}}$. We now take the common terms of multiplication on both sides and compare the remaining terms. Now we make addition and subtraction operations on both sides and make the calculations required to get the required result for ‘n’.

Complete step by step answer:
We have given that ${}^{n}{{C}_{3}}+{}^{n}{{C}_{4}}>{}^{n+1}{{C}_{3}}$, and we need to find the value of ‘n’.
$\Rightarrow $ ${}^{n}{{C}_{3}}+{}^{n}{{C}_{4}}>{}^{n+1}{{C}_{3}}$.
We know that the value of ${}^{n}{{C}_{r}}$ is given as ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
$\Rightarrow $ $\dfrac{n!}{3!\left( n-3 \right)!}+\dfrac{n!}{4!\left( n-4 \right)!}>\dfrac{\left( n+1 \right)!}{3!\left( n+1-3 \right)!}$.
$\Rightarrow $ $\dfrac{n!}{3!\left( n-3 \right)!}+\dfrac{n!}{4!\left( n-4 \right)!}>\dfrac{\left( n+1 \right)!}{3!\left( n-2 \right)!}$.
We know that a! is defined as $a!=a.\left( a-1 \right).\left( a-2 \right)......3.2.1$ and the value of (n-2) is greater than the value of (n-3) for any value of n $\left( n>o \right)$. Also, the value of (n-4) is less than the value of (n-3) and (n-2) for any value of n $\left( n>o \right)$. We also know that $a!=a\times \left( a-1 \right)!=a\times \left( a-1 \right)\times \left( a-2 \right)!$.
$\Rightarrow $ $\dfrac{n!}{3!\left( n-4 \right)!}\left( \dfrac{1}{\left( n-3 \right)}+\dfrac{1}{4} \right)>\dfrac{n!}{3!\left( n-4 \right)!}.\left( \dfrac{\left( n+1 \right)}{\left( n-3 \right).\left( n-2 \right)} \right)$.
$\Rightarrow $ $\left( \dfrac{4+\left( n-3 \right)}{4.\left( n-3 \right)} \right)>\left( \dfrac{\left( n+1 \right)}{\left( n-3 \right).\left( n-2 \right)} \right)$.
$\Rightarrow $ $\left( \dfrac{\left( n+1 \right)}{4.\left( n-3 \right)} \right)>\left( \dfrac{\left( n+1 \right)}{\left( n-3 \right).\left( n-2 \right)} \right)$.
$\Rightarrow $ $\left( \dfrac{1}{4} \right)>\left( \dfrac{1}{\left( n-2 \right)} \right)$.
$\Rightarrow $ (n – 2) > 4.
$\Rightarrow $ n > 4 + 2.
$\Rightarrow $ n > 6.
We have found the value of interval for n as n > 6.

So, the correct answer is “Option A”.

Note: We can solve the problems alternatively as follows:
$\Rightarrow $ ${}^{n}{{C}_{3}}+{}^{n}{{C}_{4}}>{}^{n+1}{{C}_{3}}$ -(1).
We use the result ${}^{n}{{C}_{r-1}}+{}^{n}{{C}_{r}}={}^{n+1}{{C}_{r}}$ in equation (1).
$\Rightarrow $ ${}^{n+1}{{C}_{4}}>{}^{n+1}{{C}_{3}}$. Now, we use ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
$\Rightarrow $ $\dfrac{\left( n+1 \right)!}{4!\left( n+1-4 \right)!}>\dfrac{\left( n+1 \right)!}{3!\left( n+1-3 \right)!}$.
$\Rightarrow $ $\dfrac{\left( n+1 \right)!}{4!\left( n-3 \right)!}>\dfrac{\left( n+1 \right)!}{3!\left( n-2 \right)!}$.
$\Rightarrow $ $\dfrac{1}{4}>\dfrac{1}{\left( n-2 \right)}$.
$\Rightarrow $ (n – 2) > 4.
$\Rightarrow $ n > 4 + 2.
$\Rightarrow $ n > 6.
We have found the value of interval for n as n > 6.