Question

In how many ways can $5$ members forming a committee out of $10$ be selected so that two particular members must be included.

Answer 1

${5_C}_2$

Well, you choose 2 members out of the 2 which must be selected, there’s only 1 way to do that.

Then you choose 3 more members out of the 8 who remain. There are ${8_C}_3$ ways to do that, which is 8!/(3!*5!)= 56ways