Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?
share|cite|improve this question
1. If we are choosing President, VP, then $156$ is fine. If we are choosing two co-presidents, I would say $\binom{13}{2}$. 2. Good. 3. Hard to know, very poor wording, what does first one mean? If we decide it just means at least one boy, tjem we wamt No. of undrestricted choices minus number of all girl choices, $\binom{13}{2}-\binom{5}{2}=68. But the wording very unclear. – André Nicolas Oct 16 '11 at 4:08
@AndréNicolas: In the book the first answer comes out to be 78 – Fahad Uddin Oct 16 '11 at 4:13
It means they interpreted as committee of two, and this is the $\binom{13}{2}$ that I mentioned. – André Nicolas Oct 16 '11 at 4:16
@AndréNicolas: How would I come to know if it has to be done like this? I have seen similar problems implemented with the way I did it.For example, you have 5 seats are 4 people are to be seated then there would be ways to seat them. – Fahad Uddin Oct 16 '11 at 4:32
@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. – Andrea Mori Oct 16 '11 at 7:46
up vote 3 down vote accepted

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)?

share|cite|improve this answer

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.

share|cite|improve this answer
Thanks for helping out :) – Fahad Uddin Oct 16 '11 at 13:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.