A) 12
B) 10
C) 8
D) 6
Correct Answer: A
Solution :
[a] \[^{x+2}{{P}_{x+2}}=a\Rightarrow a=(x+2)!\] \[^{x}{{P}_{11}}=b\Rightarrow b=\frac{x!}{(x-11)!}s\] and \[^{x-11}{{P}_{x-11}}=c\] \[\Rightarrow c=(x-11)!;\,\,\,a=182bc\] \[\therefore (x+2)!=182\frac{x!}{(x-11)!}(x-11)!\] \[\Rightarrow (x+2)(x+1)=182=14\times 13\] \[\therefore x+1=13\] \[\therefore x=12\]
You need to login to perform this action.
You will be redirected in
3 sec
