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Question: If we have given \({}^{n}{{C}_{3}}+{}^{n}{{C}_{4}}>{}^{n+1}{{C}_{3}}\), then which of the following ...

If we have given nC3+nC4>n+1C3{}^{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.

Explanation

Solution

We start solving the problem by substituting nCr=n!r!(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} in place of nC3{}^{n}{{C}_{3}}, nC4{}^{n}{{C}_{4}} and n+1C3{}^{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 nC3+nC4>n+1C3{}^{n}{{C}_{3}}+{}^{n}{{C}_{4}}>{}^{n+1}{{C}_{3}}, and we need to find the value of ‘n’.
⇒\Rightarrow nC3+nC4>n+1C3{}^{n}{{C}_{3}}+{}^{n}{{C}_{4}}>{}^{n+1}{{C}_{3}}.
We know that the value of nCr{}^{n}{{C}_{r}} is given as nCr=n!r!(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}.
⇒\Rightarrow n!3!(n−3)!+n!4!(n−4)!>(n+1)!3!(n+1−3)!\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 n!3!(n−3)!+n!4!(n−4)!>(n+1)!3!(n−2)!\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.(a−1).(a−2)......3.2.1a!=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 (n>o)\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 (n>o)\left( n>o \right). We also know that a!=a×(a−1)!=a×(a−1)×(a−2)!a!=a\times \left( a-1 \right)!=a\times \left( a-1 \right)\times \left( a-2 \right)!.
⇒\Rightarrow n!3!(n−4)!(1(n−3)+14)>n!3!(n−4)!.((n+1)(n−3).(n−2))\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 (4+(n−3)4.(n−3))>((n+1)(n−3).(n−2))\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 ((n+1)4.(n−3))>((n+1)(n−3).(n−2))\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 (14)>(1(n−2))\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 nC3+nC4>n+1C3{}^{n}{{C}_{3}}+{}^{n}{{C}_{4}}>{}^{n+1}{{C}_{3}} -(1).
We use the result nCr−1+nCr=n+1Cr{}^{n}{{C}_{r-1}}+{}^{n}{{C}_{r}}={}^{n+1}{{C}_{r}} in equation (1).
⇒\Rightarrow n+1C4>n+1C3{}^{n+1}{{C}_{4}}>{}^{n+1}{{C}_{3}}. Now, we use nCr=n!r!(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}.
⇒\Rightarrow (n+1)!4!(n+1−4)!>(n+1)!3!(n+1−3)!\dfrac{\left( n+1 \right)!}{4!\left( n+1-4 \right)!}>\dfrac{\left( n+1 \right)!}{3!\left( n+1-3 \right)!}.
⇒\Rightarrow (n+1)!4!(n−3)!>(n+1)!3!(n−2)!\dfrac{\left( n+1 \right)!}{4!\left( n-3 \right)!}>\dfrac{\left( n+1 \right)!}{3!\left( n-2 \right)!}.
⇒\Rightarrow 14>1(n−2)\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.