if 2 students are in committee, then we have to choose 3 people from 6 parents and 6 village leaders. So, number of ways to choose 2 students in the committee:
"n=C^3_{12}=\\frac{12!}{9!3!}=220"
total number of ways to choose 5 people in the committee:
"N=C^5_{17}=\\frac{17!}{12!5!}=6188"
the probability that 2 students are in the committee:
"P=\\frac{n}{N}=\\frac{220}{6188}=\\frac{55}{1547}"
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