Q:

Sample Points Calculation.

A coin is tossed three times. There are three sample points that i can get one head and two tail.
I can count the sample points after writing the sample space, as , 
(H, H, H), (H, H, T), (H, T, H), (H, T, T), 
(T, H, H), (T, H, T), (T, T, H) and (T, T, T)
Is there any way to count that there are three sample points of getting one head and two tail without writing the sample space ?
EDIT:
for example : A committee of three is chosen from five councilors - Adams, Burke, Cobb, Dilby and Evans.
What is the probability Burke is on the committee?
solution : 
Abbreviate the names of the five councilors with the letters A, B, C, D and E.
There are 10 possible committees: 
(A, B, C), (A, B, D), (A, B, E), (A, C, D), (A, C, E), (A, D, E), (B, C, D), (B, C, E), (B, D, E) and (C, D, E)
Of these, Burke is included in 6:
(A, B, C), (A, B, D), (A, B, E), (B, C, D), (B, C, E) and (B, D, E)
So
The Number of ways it can happen = 6
The Total number of outcomes = $\binom{5}{3}$
But i don't want to write down the sample points where Bruke is included. How can i know that ?

A:

Yes.  Think of three slots being filled with either an H or a T.  How many ways are there to pick the one slot that will be filled with an H?  This is solved with Combinations.$${3\choose1}=3$$
For your committee problem, work it this way.  How many total committees are there?$${5\choose3}=10$$  How many committees have Brandon is asking how many ways can you make a two person committee from 4 people(Brandon plus two other people, chosen from the 4 remaining candidates)?$${4\choose2}=6$$Therefore $\frac{6}{10}$ of the committees have Brandon.

