Prove that


Find $n$, If ${ }^{n+5} P_{n+1}=\frac{11}{2}(n-1) \cdot{ }^{n+3} P_{n}$, 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,

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

${ }^{n+5} P_{n+1}=\frac{11}{2}(n-1) \cdot{ }^{n+3} P_{n}$

$\frac{(n+5) !}{4 !}=\frac{11}{2}(n-1) \frac{(n+3) !}{3 !}$

$\frac{(n+5)(n+4)(n+3) !}{4 \times 3 !}=\frac{11}{2}(n-1) \frac{(n+3) !}{3 !}$


$n^{2}+9 n+20=22 n-22$

$n^{2}-13 n+42=0$



Hence, values of n are 7 & 6


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