If ${ }^{2 n-1} P_{n}:{ }^{2 n+1} P_{n-1},=22: 7$, find $n .$
To find: the value of n
Formula Used:
Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,
${ }_{n p_{r}}=\frac{n !}{(n-r) !}$
${ }^{2 n-1} P_{n}:{ }^{2 n+1} P_{n-1},=22: 7$
$\frac{(2 n-1) !}{(n-1) !}: \frac{(2 n+1) !}{(n+2) !}=\frac{22}{7}$
$\frac{(2 n-1) !}{(n-1) !}: \frac{(2 n+1)(2 n)(2 n-1) !}{(n+2)(n+1) n(n-1) !}=\frac{22}{7}$
$\frac{(2 n-1) !}{(n-1) !} \times \frac{(n+2)(n+1) n(n-1) !}{(2 n+1)(2 n)(2 n-1) !}=\frac{22}{7}$
$\frac{(n+2)(n+1)}{(2 n+1) 2}=\frac{22}{7}$
$\frac{n^{2}+3 n+2}{2 n+1}=\frac{44}{7}$
$7 n^{2}+21 n+14=88 n+44$
$7 n^{2}-67 n-30=0$
$n=10,-0.42$
Since $\mathrm{n}$ cannot be $-0.42$
Hence, value of $n$ is 10 .