Question

If 6!(1/2!+4/3!)=7pr.find r?

Answer 1

The value of r is 4.

• 6!($\frac{1}{2!}$+$\frac{4}{3!}$) = ${}^7P_r$

• We have to find the value of r.

• ${}^nP_r$ is represented as $\frac{n!}{(n-r)!}$.

• Now, ${}^7P_r$ = $\frac{7!}{(7-r)!}$

• We now solve the LHS of the equation,

LHS = 6!($\frac{1}{2!}$+$\frac{4}{3!}$)

LHS = 6!($\frac{1}{2}$+$\frac{4}{6}$)   (as 2!= 2 and 3!=6)

LHS = 6!($\frac{3+4}{6}$)

LHS = 6!($\frac{7}{6}$)

LHS = $\frac{6! . 7}{6}$

LHS = $\frac{7!}{6}$               (n × (n-1)! = n!)

• Now equating both sides, we get

• (7-r)! = 6    

• As we know that 3! =6,

• Therefore (7-r) = 3

r = 4