A) 15
B) 12
C) 10
D) 18
Correct Answer: B
Solution :
We have \[a{{=}^{x+2}}{{P}_{x+2}}=(x+2)\ !,\ b{{=}^{x}}{{P}_{11}}=\frac{x\ !}{(x-11)\ !}\] and \[c{{=}^{x-11}}{{P}_{x-11}}=(x-11)\ !\] Now \[a=182\ bc\Rightarrow (x+2)\ !=182\ .\ \frac{x\ !}{(x-11)\ !}(x-11)\ !\] \[\Rightarrow \ (x+2)\,!\ =182x\ !\Rightarrow (x+2)(x+1)=182\]\[\Rightarrow \]\[x=12\].You need to login to perform this action.
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