Question

If ${ }^{2 n+1} P _{n-1}:{ }^{2 n-1} P _n=11: 21$, then $n^2+n+15$ is equal to :

Hint:

When solving permutation ratio problems, simplify the expressions carefully and solve for \( n \). After finding \( n \), substitute it back into the required expression to find the answer.

Answer 1

Correct Answer: 45

$\frac{(2n+1)!(n-1)!}{(n+2)!(2n-1)!}=\frac{11}{21}$
$\Rightarrow \frac{(2n+1)(2n)}{(n+2)(n+1)n}=\frac{11}{21}$
$\Rightarrow \frac{2n+1}{(n+1)(n+2)}=\frac{11}{42}$
$\Rightarrow n=5$
$\Rightarrow n^2+n+15$
$=25+5+15=45$
So , the correct answer is 45.

Answer 2

We are given the following equation: \[ \frac{(2n+1)P_{n-1}}{2nP_n} = \frac{11}{21}. \] Step 1: We know that the permutation formula is \( nP_r = \frac{n!}{(n-r)!} \). Therefore: \[ (2n+1)P_{n-1} = \frac{(2n+1)!}{(2n+1-(n-1))!} = \frac{(2n+1)!}{n!}, \] \[ 2nP_n = \frac{(2n)!}{(2n-n)!} = \frac{(2n)!}{n!}. \] Step 2: Now, substitute these expressions into the given equation: \[ \frac{\frac{(2n+1)!}{n!}}{\frac{(2n)!}{n!}} = \frac{11}{21}. \] Simplifying: \[ \frac{(2n+1)!}{(2n)!} = \frac{11}{21}. \] \[ \frac{(2n+1)(2n)!}{(2n)!} = \frac{11}{21} \quad \Rightarrow \quad (2n+1) = \frac{11}{21}. \] Step 3: Solving for \( n \), we get: \[ 2n + 1 = 5 \quad \Rightarrow \quad 2n = 4 \quad \Rightarrow \quad n = 5. \] Step 4: Now, substitute \( n = 5 \) into the expression \( n^2 + n + 15 \): \[ n^2 + n + 15 = 5^2 + 5 + 15 = 25 + 5 + 15 = 45. \] Thus, \( n^2 + n + 15 = 45 \).