A class of 20 students has 55% girls. The top 3 students are felicitated with gold, silver and bronze medals.

Question

In how many ways would the medals be distributed if there are atleast $2$ girls among top $3$?

Will the answer be: $$3!\times\left(\binom {11}{3} + 9\times\binom{11}{2}\right)= 3960$$ or: $$3!\times\left(\binom {11}{2}\times\binom{18}{1}\right) = 5940$$(At least $2$ girls and any other person)

Answer 1

There are $3!\times\binom {11}{3}$ ways to give them all to girls. To give exactly two, choose a boy and two girls in one of $9\times\binom{11}{2}$ ways, and permute the medals. So the answer is $6(165+9\times 55)=3960$.

Answer 2

There are $4$ cases:

• girls receive gold, silver and bronze.
• girls receive gold and silver and no bronze.
• girls receive gold, bronze and no silver.
• girls receive silver, bronze and no gold.

For the first mentioned we find $11\times10\times9=990$ possibilities.

For the others coincidently the same: $11\times10$ possibilities for the girls multiplied with $9$ possibilities for the boys.

So there are $4\times990=3960$ possibilities.

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The second answer is not correct. See my comment on your question.