Question
The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by
A.
$$7!\,\, \times \,\,5!$$
B.
$$6!\,\, \times \,\,5!$$
C.
$$30!$$
D.
$$5!\,\, \times \,\,4!$$
Answer :
$$7!\,\, \times \,\,5!$$
Solution :
No. of ways in which 6 men can be arranged at a round table $$= (6 - 1)! = 5!$$
Now women can be arranged in $$^6{P_5} = 6!$$ Ways.
Total Number of ways = $$6!\,\, \times \,\,5!$$