This is a pairing of two classic types of the same problem.
Problem A has \(100\) possible outcomes: \(10\times 10\) ordered pairs.
Problem B has \(\dbinom{10}{2}=45\) possible outcomes: that is choosing two from ten.
PART A: Of the one hundred ordered pairs, there are \(19\) pairs that contain Colin at least once.
PART B: Of the forty-five two-selections, Colin is in nine of the selections.
Clay, you need to use the above and post your results. Please, please do it.