As you have stated the problem, the probability is 1, because since there are 7 males, there are of course 6 males, and since there are 5 females there are of course 3 females. So you left out the word "particular". Your question should be: What is the probability that 6 PARTICULAR male and 3 PARTICULAR female students ARE included inthe meeting. So the 20 females consist of 5 particular females, and 15 non-particular females. And, the 30 males consist of 6 particular males, and 24 non-particular males. The number of successful choices are 1. Choose all 3 particular females 1 way and 2 non-particular females C(15,2)=105 ways. 2. Choose all 6 particular males 1 way, and 1 non-particular male C(24,1)=24 ways. That's (105)(24) = 2520 successful ways. The number of possible choices are 1. Choose any 5 females C(20,5)=15504 ways. 2. Choose any 7 males C(30,7)=2035800 ways ways. That's (15504)(2035800) = 31563043200 possible ways. So the probability is 2520/31563043200 which reduces to 7/87675120 or about 0.00000008 Edwin