In a group of $10$ children no two kids have the same name. We know that Mary and Jane are among these children. In how many ways can we select $5$ children from this group so that Mary and Jane are always in the selection?
So I started by looking at all the possible outcomes
$$\frac{n!}{r!(n-r)!} \implies \frac{10!}{5!(10-5)!} = 6048$$
from here I get stuck and I don't know how to proceed. Thanks for your help in advance.
If Mary and Jane are always in the selection, then $3$ more children have to be selected from the remaining $8$ children. Number of ways of selecting $3$ children from $8$ children is
$$= \frac{8!}{(8-3)! \cdot 3!} = 56$$