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Permutations and Combinations question

2023 · 8 Apr · Shift 1 · Q27
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Permutations and Combinations question

2023 · 8 Apr · Shift 1 · Q27

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
The number of ways, in which 5 girls and 7 boys can be seated at a round table so that no two girls sit together, is :
  1. A
    720
  2. B
    7(360)27(360)^{2}7(360)2
  3. C
    7(720)27(720)^{2}7(720)2
  4. D
    126(5!)2126(5 !)^{2}126(5!)2
View written solutionFree

Correct answer: D

  1. Arrange the boys first around the round table

Since seating is around a round table, arrangements differing only by rotation are the same.

So, the number of circular arrangements of 777 boys is

(7−1)!=6!=720.(7-1)! = 6! = 720.(7−1)!=6!=720.
  1. Identify places for the girls

After arranging the 777 boys, there are exactly 777 gaps between consecutive boys:

_ B _ B _ B _ B _ B _ B _ B\_\,B\,\_\,B\,\_\,B\,\_\,B\,\_\,B\,\_\,B\,\_\,B_B_B_B_B_B_B_B

To ensure that no two girls sit together, each girl must occupy a different gap.

We need to choose 555 of these 777 gaps for the 555 girls:

(75).\binom{7}{5}.(57​).
  1. Arrange the girls in the chosen gaps

The 555 girls can be arranged in those selected 555 gaps in

5!5!5!

ways.

  1. Total number of seatings

Thus total number of ways is

6!⋅(75)⋅5!.6! \cdot \binom{7}{5} \cdot 5!.6!⋅(57​)⋅5!.

Now,

(75)=21,\binom{7}{5} = 21,(57​)=21,

so

6!⋅21⋅5!=720⋅21⋅120.6! \cdot 21 \cdot 5! = 720 \cdot 21 \cdot 120.6!⋅21⋅5!=720⋅21⋅120.

Also,

21=7⋅62,21 = \frac{7\cdot 6}{2},21=27⋅6​,

so this can be written as

720⋅21⋅120=126⋅(120)2=126(5!)2.720 \cdot 21 \cdot 120 = 126 \cdot (120)^2 = 126(5!)^2.720⋅21⋅120=126⋅(120)2=126(5!)2.
  1. Match with the options

Since 5!=1205! = 1205!=120, we get

126(5!)2.126(5!)^2.126(5!)2.

This is Option D.

Verification with stored answer: Stored correct answer is D, which matches our result.

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