Permutations & Combinations Question - None

Question

If $^{2n+1}P_{n-1} : ^{2n-1}P_n = 3 : 5$, then the value of $n$ is equal to

Accepted Answer

Step 1: We have $^{2n+1}P_{n-1} : ^{2n-1}P_n = 3 : 5$

Step 2: $\frac{^{2n+1}P_{n-1}}{^{2n-1}P_n} = \frac{3}{5}$

Step 3: $\frac{\frac{(2n+1)!}{(n+2)!}}{\frac{(2n-1)!}{(n-1)!}} = \frac{3}{5}$

Step 4: $\frac{(2n+1)! \cdot (n-1)!}{(n+2)! \cdot (2n-1)!} = \frac{3}{5}$

Step 5: $\frac{(2n+1)(2n) \cdot n!}{(n+2)(n+1)n \cdot 2n!} = \frac{3}{5}$

Step 6: $\frac{(2n+1)(2n)}{(n+2)(n+1)n} = \frac{3}{5}$

Step 7: $10(2n+1) = 3(n+2)(n+1)$

Step 8: Solving this equation gives $n = 4$

∴ Answer is (b).

Question: