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Permutations and Combinations question

2019 · 9 Jan · Shift 1 · Q40
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Permutations and Combinations question

2019 · 9 Jan · Shift 1 · Q40

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is :
  1. A
    500
  2. B
    350
  3. C
    200
  4. D
    300
View written solutionFree

Correct answer: D

  1. Total number of teams without restriction

We need a team of 2 girls and 3 boys from a class of 5 girls and 7 boys.

  • Number of ways to choose 2 girls from 5: (52)=10\binom{5}{2} = 10(25​)=10
  • Number of ways to choose 3 boys from 7: (73)=35\binom{7}{3} = 35(37​)=35

So total teams without any restriction: (52)(73)=10×35=350\binom{5}{2}\binom{7}{3} = 10 \times 35 = 350(25​)(37​)=10×35=350

  1. Count teams where boys A and B are together

Since A and B refuse to be on the same team, we must exclude all teams containing both A and B.

If A and B are both selected, then among the 3 boys in the team, the third boy must be chosen from the remaining 5 boys: (51)=5\binom{5}{1} = 5(15​)=5

The 2 girls can still be chosen from 5 girls in: (52)=10\binom{5}{2} = 10(25​)=10

Hence, number of teams containing both A and B: (52)(51)=10×5=50\binom{5}{2}\binom{5}{1} = 10 \times 5 = 50(25​)(15​)=10×5=50

  1. Required number of teams

Subtract the invalid teams from the total: 350−50=300350 - 50 = 300350−50=300

  1. Check options
  • A: 500 ❌
  • B: 350 ❌
  • C: 200 ❌
  • D: 300 ✅

Therefore, the correct answer is: 300\boxed{300}300​

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