Question

The no. of ways in which 6 gentlemen and 3 ladies be seated round a table so that every gentleman may have a lady by his side is ...

• A. 1440
• B. 720
• C. 240
• D. 480

Answer 1

Answer: (A)

1440

First three ladies can be seated in $(3-1)!=2!$ ways.

In the intermediate gaps between ladies two persons are to be placed.

$\therefore$ Number of groupings = $\frac{6!}{2!2!2!}$

but as each person is different, no of permutations $=\frac{6!}{2!2!2!}\times 2!\times 2!\times 2!$

$=6!$

$\therefore$ Total no. of ways $=6!\times 2=720\times 2=1440$