Solve the following


If ${ }^{k+5} P_{k+1}=\frac{11(k-1)}{2} \cdot{ }^{k+3} P_{k}$, then the values of $k$ are

(a) 7 and 11

(b) 6 and 7

(c) 2 and 11

(d) 2 and 6


(b) 6 and 7

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

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


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

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



$\Rightarrow k^{2}+9 k+20=22 k-22$

$\Rightarrow k^{2}-13 k+42=0$


$\Rightarrow k=6,7$

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