To find out how many combinations of N objects taken either A or B at time, add both of the individual combinations
_{N}C
_{a} +
_{N}C
_{b}
Applied Example
How many ways can four students in a class of ten be chosen to form a 'math appreciation' committee?
 _{10}C_{4} = 10!/(4! × 6!) = 210
How many way can three students in that same class be chosen to form the math committee?
 _{10}C_{3} = 10!/(3! × 7!) = 120
How many ways can four or three students in that class of 10 be chosen to form this committee?

(_{10}C_{4}) + (_{10}C_{3}) = 210 + 120 = 330 different ways
Practice Problems
Problem 1) There are eight candidates at school running for three seats in the student government. If you can vote for three or few students. In how many different ways can you vote?
Answer
_{8}C_{3}+ _{8}C_{2} + _{8}C_{1} =56 +28 +8 =
92
Problem 2) Fourteen students want to be in the math club, one of the most popular clubs at school. The club will be able to admit four or fewer students (Only students with at least an A average in math will be admitted). In how many different combinatios can students join the club?
Answer
_{14}C_{4}+ _{14}C_{3} + _{14}C_{2} + _{14}C_{1} = 1470