Q56 / 150 · MHT CET 2026 (19 April Shift 1) · Mathematics
Four men and three women are to be arranged at a round table. The number of arrangements in which no two women are together is ....
- A$7!$
- Correct answerB$3! \times 4!$
- C$3! \times 3!$
- D$6!$
Solution
First, arrange the $4$ men around the circular table. The number of ways to do this is $(4 - 1)! = 3!$.
When the $4$ men are seated, they create $4$ gaps between them.
To ensure that no two women sit together, the $3$ women must be placed in these $4$ gaps.
The number of ways to choose $3$ gaps out of $4$ and arrange the $3$ women in them is $^{4}P_{3} = 4 \times 3 \times 2 = 4!$.
Therefore, the total number of arrangements is $3! \times 4!$.
Answer: $3! \times 4!$
When the $4$ men are seated, they create $4$ gaps between them.
To ensure that no two women sit together, the $3$ women must be placed in these $4$ gaps.
The number of ways to choose $3$ gaps out of $4$ and arrange the $3$ women in them is $^{4}P_{3} = 4 \times 3 \times 2 = 4!$.
Therefore, the total number of arrangements is $3! \times 4!$.
Answer: $3! \times 4!$
Hence, the correct answer is option: 'B'.