Solve this


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,

${ }_{n p_{r}}=\frac{n !}{(n-r) !}$

${ }^{2 n-1} P_{n}:{ }^{2 n+1} P_{n-1},=22: 7$

$\frac{(2 n-1) !}{(n-1) !}: \frac{(2 n+1) !}{(n+2) !}=\frac{22}{7}$

$\frac{(2 n-1) !}{(n-1) !}: \frac{(2 n+1)(2 n)(2 n-1) !}{(n+2)(n+1) n(n-1) !}=\frac{22}{7}$

$\frac{(2 n-1) !}{(n-1) !} \times \frac{(n+2)(n+1) n(n-1) !}{(2 n+1)(2 n)(2 n-1) !}=\frac{22}{7}$

$\frac{(n+2)(n+1)}{(2 n+1) 2}=\frac{22}{7}$

$\frac{n^{2}+3 n+2}{2 n+1}=\frac{44}{7}$

$7 n^{2}+21 n+14=88 n+44$

$7 n^{2}-67 n-30=0$


Since $\mathrm{n}$ cannot be $-0.42$

Hence, value of $n$ is 10 . 

Leave a comment