Question

Find the value of \[n\] for which \[{}^{n - 1}{C_4} - {}^{n - 1}{C_3} - \dfrac{5}{4} \times {}^{n - 2}{P_2} < 0\] , where \[n \in N\].
A. \[\left\{ {5,6,7,8,9,10} \right\}\]
B. \[\left\{ {1,2,3,4,5,6,7,8,9,10} \right\}\]
C. \[\left\{ {1,4,5,6,7,8,9,10} \right\}\]
D. \[\left( { - \infty ,2} \right) \cup \left( {3,11} \right)\]

Answer 1

Hint: Here, one inequality equation is given. First, using the formulas of permutation and combination simplify the inequality equation. After that, solve the inequality equation by applying basic mathematical operations. In the end, check the values of \[n\] with the inequality equations and get the required answer.

Formula Used: Permutation Formula: \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Combination Formula: \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]

Complete step by step solution: The given inequality equation is \[{}^{n - 1}{C_4} - {}^{n - 1}{C_3} - \dfrac{5}{4} \times {}^{n - 2}{P_2} < 0\].
Let’s simplify the inequality equation by applying the permutation and combination formulas.
We get,
\[\dfrac{{\left( {n - 1} \right)!}}{{4!\left( {n - 5} \right)!}} - \dfrac{{\left( {n - 1} \right)!}}{{3!\left( {n - 4} \right)!}} - \dfrac{5}{4} \times \dfrac{{\left( {n - 2} \right)!}}{{\left( {n - 4} \right)!}} < 0\]
Simplify the numerator and denominator by applying the factorial.
\[\dfrac{{\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)\left( {n - 5} \right)!}}{{4!\left( {n - 5} \right)!}} - \dfrac{{\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)!}}{{3!\left( {n - 4} \right)!}} - \dfrac{5}{4} \times \dfrac{{\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)!}}{{\left( {n - 4} \right)!}} < 0\]
Cancel out the common terms from the numerator and the denominator.
\[\dfrac{{\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)}}{{4!}} - \dfrac{{\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{3!}} - \dfrac{5}{4}\left( {n - 2} \right)\left( {n - 3} \right) < 0\]
Multiply both sides by \[4!\].
 \[\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right) - 4\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right) - 30\left( {n - 2} \right)\left( {n - 3} \right) < 0\]
\[ \Rightarrow \left( {n - 2} \right)\left( {n - 3} \right)\left[ {\left( {n - 1} \right)\left( {n - 4} \right) - 4\left( {n - 1} \right) - 30} \right] < 0\]
\[ \Rightarrow \left( {n - 2} \right)\left( {n - 3} \right)\left[ {{n^2} - 5n + 4 - 4n + 4 - 30} \right] < 0\]
\[ \Rightarrow \left( {n - 2} \right)\left( {n - 3} \right)\left[ {{n^2} - 9n - 22} \right] < 0\]
\[ \Rightarrow \left( {n - 2} \right)\left( {n - 3} \right)\left( {n + 2} \right)\left( {n - 11} \right) < 0\]
Solving the above inequality, we get
\[n < 2\] , \[n < 3\], \[n < - 2\], and \[n < 11\]
\[ \Rightarrow - 2 < n < 2\] or \[3 < n < 11\]
\[ \Rightarrow \left( { - \infty ,2} \right) \cup \left( {3,11} \right)\]
\[ \Rightarrow \left( {0,2} \right) \cup \left( {3,11} \right)\]
But it is given that \[n \in N\]
So, \[n = 1,4,5,6,7,8,9,10\] \[.....\left( 1 \right)\]

Also, from the given inequality equation, we get
\[n - 1 \ge 4\], \[n - 1 \ge 3\], and \[n - 2 \ge 2\]
\[ \Rightarrow n \ge 5\], \[n \ge 4\], and \[n \ge 4\]
So, \[n \ge 5\] \[.....\left( 2 \right)\]

From the equations \[\left( 1 \right)\] and \[\left( 2 \right)\], we get
\[n = 5,6,7,8,9,10\]
Therefore, the values of \[n\] are \[\left\{ {5,6,7,8,9,10} \right\}\].

Option ‘A’ is correct

Note: Students directly solve the factorial equations without simplifying or cancelling it.
Factorial of a number is a product of all whole numbers less than that number up to 1.
It is defined as: \[n! = n\left( {n - 1} \right)\left( {n - 2} \right)....3 \times 2 \times 1\]