Question

$7$ gentlemen and $3$ ladies are to be seated in a row. If the ladies insist on sitting together while two of the gentlemen refuse to take consecutive seats, then the number of ways in which they can be seated is

• A. $2399976$
• B. $21844$
• C. $630624$
• D. $181440$

Answer 1

The correct option is D $181440$

Total number of ways $=$ number of ways in which three ladies are together $-$ number of ways in which three ladies are together and those two gentlemen are together.
Number of ways in which three ladies are together $=8!\times 3!$
(consider three ladies as one unit and internal arrangement of ladies)
Number of ways in which three ladies are together and those two gentlemen are together $=3!\times 2!\times 7!$
(consider three ladies as one unit and two gentlemen as one unit and internal arrangement of ladies and gentlemen)
Total number of required ways $=3!8!-3!2!7!=181440$

Alternate method:
$10$ Guest $=3L+7G$
Let two particular gentle-men who refused to sit together be $G_1,G_2.$
Total arrangement of $5$ other gentle-men considering $3$ ladies as $1$ string
$=6!3!$
Now, $G_1,G_2$ can be arranged in gaps of above arrangement in ${}^7{P}_2$ ways
$\therefore$ Required number of ways $=6!3!{}^7{P}_2$