Mrs. Claus is selecting three reindeer from the herd for a special

Mrs. Claus is selecting three reindeer from the herd for a special sleigh ride. Among the herd are Rudolph, Blitzen, and Comet. What is the probability that she chooses Rudolph, but not Blitzen and not Comet?

It really doesn't matter who is being chosen and who is being left. For the probability we require
a) Ways 3 reindeer can be chosen from n, in which R is selected but B and C are not selected. => one seat taken by R, the remaining two seats have to be selected from n-3(R, B and C) => n-3C2
b) Total ways of selecting 3 out of n, that is nC3, where n is total reindeers.
\(P = \frac{n-3C2}{nC3}\)
So, what you have to find is n.

(1) The probability that she chooses Rudolph and Blitzen, but not Comet, is 1/14.
Choosing two and not the third gives a probability as \(\frac{n-3C1}{nC3} = \frac{1}{14}\)
How?? 
a) Ways 3 reindeer can be chosen from n, in which R is not selected but B and C are selected. => one seat each taken by B and C, the remaining one seat has to be selected from n-3(R, B and C) => n-3C1
b) Total ways of selecting 3 out of n, that is nC3, where n is total reindeers.
\(P = \frac{n-3C1}{nC3}\)​​​​​​​
Only one variable, and you will get the answer.
Sufficient

(2) The probability that she chooses none of Rudolph, Blitzen, and Comet is 5/21.
Choosing none of the three means selecting three from remaining n-3 and gives a probability of \(\frac{n-3C3}{nC3} = \frac{5}{21}\)
Only one variable, and you will get the answer.
Sufficient

When you solve, which is not required here, n is 9.

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