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Permutations and Combinations question

2003 · Shift 0 · Q107
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Permutations and Combinations question

2003 · Shift 0 · Q107

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
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
  1. A
    7! ×5!  7!\, \times 5!\,\,7!×5!
  2. B
    6! ×5!6!\, \times 5!6!×5!
  3. C
    30!30!30!
  4. D
    5! ×4!5!\, \times 4!5!×4!
View written solutionFree

Correct answer: B

  1. Arrange the 6 men first around the round table

For circular arrangements of nnn distinct people, the number of ways is (n−1)!(n-1)!(n−1)!.

So, the number of ways to arrange 6 men around the table is

(6−1)!=5!(6-1)! = 5!(6−1)!=5!
  1. Identify the gaps for women

Once the 6 men are seated, there are exactly 6 gaps between consecutive men around the circle:

_ M _ M _ M _ M _ M _ M\_\ M\ \_\ M\ \_\ M\ \_\ M\ \_\ M\ \_\ M_ M _ M _ M _ M _ M _ M

To ensure that no two women sit together, each woman must occupy a different gap.

  1. Choose gaps for the 5 women

Out of the 6 available gaps, we must choose 5 gaps:

(65)=6\binom{6}{5} = 6(56​)=6
  1. Arrange the 5 women in those chosen gaps

The 5 women can be arranged in these 5 selected gaps in

5!5!5!

ways.

  1. Total number of arrangements

Thus total number of valid seatings is

5!×(65)×5!=5!×6×5!5! \times \binom{6}{5} \times 5! = 5! \times 6 \times 5!5!×(56​)×5!=5!×6×5!

Since

6×5!=6!6 \times 5! = 6!6×5!=6!

we get

5!×6!=6!×5!5! \times 6! = 6! \times 5!5!×6!=6!×5!
  1. Match with the options

This is exactly Option B:

6!×5!6! \times 5!6!×5!
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