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Permutations and Combinations question

2021 · 17 Mar · Shift 1 · Q35
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Permutations and Combinations question

2021 · 17 Mar · Shift 1 · Q35

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
Team 'A' consists of 7 boys and n girls and Team 'B' has 4 boys and 6 girls. If a total of 52 single matches can be arranged between these two teams when a boy plays against a boy and a girl plays against a girl, then n is equal to :
  1. A
    5
  2. B
    2
  3. C
    4
  4. D
    6
View written solutionFree

Correct answer: C

  1. Interpret the match condition

Each single match is arranged only between players of the same gender:

  • a boy from Team A plays a boy from Team B
  • a girl from Team A plays a girl from Team B

So total matches are: (boy-boy matches)+(girl-girl matches)\text{(boy-boy matches)} + \text{(girl-girl matches)}(boy-boy matches)+(girl-girl matches)

  1. Count boy-boy matches

Team A has 777 boys and Team B has 444 boys.

Hence number of boy-boy matches: 7×4=287 \times 4 = 287×4=28

  1. Count girl-girl matches

Team A has nnn girls and Team B has 666 girls.

Hence number of girl-girl matches: n×6=6nn \times 6 = 6nn×6=6n

  1. Use the total number of matches

Given total matches = 525252: 28+6n=5228 + 6n = 5228+6n=52

  1. Solve for nnn

6n=52−28=246n = 52 - 28 = 246n=52−28=24 n=246=4n = \frac{24}{6} = 4n=624​=4

  1. Check options
  • A: 555 ❌
  • B: 222 ❌
  • C: 444 ✅
  • D: 666 ❌

Therefore, the correct answer is: 4\boxed{4}4​ which is Option C.

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