$8$ women and $6$ men are standing in a line.
- How many arrangements are possible if any individual can stand in any position?
- In how many arrangements will all $6$ men be standing next to one another?
- In how many arrangements will no two men be standing next to one another?
- There are $8$ women and $6$ men i.e. $14$ people.
$\therefore \;$ Number of arrangements possible if any individual can stand in any position $= 14!$ - Consider the $6$ men as ONE group.
Then $8$ women plus the $1$ group of 6 men can stand in a line in $9!$ ways.
Now, the $6$ men can, amongst themselves, stand in a line in $6!$ ways.
$\therefore \;$ Number of ways in which all $6$ men will be standing next to one another $= 9! \times 6!$ ways - $W \;\;\; M \;\;\; W \;\;\; M \;\;\; W \;\;\; M \;\;\; W \;\;\; M \;\;\; W \;\;\; M \;\;\; W \;\;\; M \;\;\; W \;\;\; W \;\;\; W$
When no two men are standing next to one another, there are $9$ places which can be occupied by the $6$ men.
$9$ places can be occupied by $6$ men in ${^{9}}{P}_{6}$ ways.
The $8$ women can be arranged amongst themselves in $8!$ ways.
$\therefore \;$ Number of ways in which 8 women and 6 men can be arranged so that no two men are standing next to one another $= 8! \times {^{9}}{P}_{6}$ ways