Question

$W_1, W_2$ and $W_3$ are three ticket windows in a cinema hall. 12 persons $\left(A_1, A_2, \ldots \ldots, A_{12}\right)$ are standing for the tickets in a queue of 4,3 and 5 number of persons infront of $W_1, W_2$ and $W_3$ respectively. $A_1$ and $A_2$ want to get their tickets from $W_1$ and $A_3, A_4$ and $A_5$ want to get their tickets from $W _3$. If the number of ways in which all are getting their tickets such that $A_1, A_2, A_3, A_4$ and $A_5$ are getting their tickets as early as possible is $k (7 !)$, then find the value of $\left[\frac{ k }{3}\right]$.
(Assuming each person is getting exactly one ticket at a time).
[Note: $[y]$ denotes greatest integer less than or equal to $y$.

Answer 1

$\text { Required number of ways }=\frac{7 !}{2 ! 2 ! 2 ! 3 !} \times 2 ! \times 2 ! \times 3 ! \times 2 ! \times 2 ! \times 3 ! $
$=12(7 !) \equiv k (7 !)$
$\therefore \left[\frac{ k }{3}\right]=\left[\frac{12}{3}\right]=4$