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Here is a propability question I am stuck with, In a class there are 8 boys and 5 girls. Two class representatives are to be choosen. In how many ways they can be selected if,

  1. both are choose from all?
  2. one is to be a boy and other a girl?
  3. the first one is to be a boy and the second any of the boy or girl?

Here is how I solved it,

  1. As there are 8 boys and 5 girls so, 8+5= 13. The choice for first will be 13 and for the second one it will be 12 which makes 156.
  2. For 8 boys choice is 8 and for 5 girls the choice is 5. By multiplication principle = 40
  3. if first one is boy choices are 8 and then anyone, choices are 12, by multiplication principle, 8*12= 96 The 1st answer does not match.Can anyone please tell me why?
  • 1
    @Akito: that is correct if you need to keep trace of the *ordering* of the seating. If not, you need to divide out by the 4!=24 ways to rearrange the 4 people seated.2011-10-16

2 Answers 2

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Hints: For 1, you need combinations. For 2, you can have any one of 8 boys and any of 5 girls. For 3, how many ways are there to pick a boy for the first, then how many ways are there to pick somebody else (assuming the two can't be the same person)?

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Answers to 1 and 2 have been already given. For problem 3, I believe that the simplest way to proceed is to count all choices of 2 (of which we know there are 78) and exclude all combinations made by two girls, which are $5\cdot4/2=10$. Thus the total choices which include at least one boy are 68.