Browse Questions

# If $(k^2-k)C_2=(k^2-k)C_4$ then $k$ is equal to

$\begin{array}{1 1}(A)\;1\\(B)\;3\\(C)\;4\\(D)\;\text{None of these}\end{array}$

Toolbox:
• If $nC_p=nC_q$ then $p+q=n$
We know that $nC_p=nC_q$ then $p+q=n$
$2+4=k^2-k$
$k^2-k-6=0$
(Or) $(k-3)(k+2)=0$
$\therefore$ k=3,-2 but -2 is -ve so rejected
$\therefore k=3$
Hence (B) is the correct answer.