Question

$W_1,W_2$ and $W_3$ are three ticket windows in a cinema hall. 12 persons $\left(A_1,A_2,\dots \dots ,A_{12}\right)$ 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

Answer: 4

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