This is a simple exercise. To solve it, we just need to apply the formula to calculate the number of r-combinations.
“The number of r-combinations of a set with n elements, where n is a nonnegative integer and r is an integer with 0 ≤ r ≤ n, equals C(n,r) = n! / r!(n−r)!” Discrete mathematics and its applications by Rosen.
Table of Contents
a) C(5,1)
C(5,1) = 5!/1!(5-1)!
= 5!/4!
= 5×4!/4!
= 5
b) C(5,3)
C(5,3)=5!/3!(5-3)!
= 5!/3!2!
=5x4x3!/3!2!
=20/2!
=10
c) C(8,4)
C(8,4)=8!/4!(8-4)!
=8x7x6x5x4!/4!4!
=1680/4!
=1680/24
=70
d) C(8,8)
C(8,8)=8!/8!(8-8)!
=1/0!
=1
e) C(8,0)
C(8,0)=8!/0!(8-0)!
= 8!/1×8!
=1
f) C(12,6)
C(12,6) = 12!/6!(12-6)!
=12x11x10x9x8x7x6!/6!6!
=665 280/ 6!
=665 280/ 720
=924
Related exercises:
- A group contains n men and n women. How many ways are there to arrange these people in a row if the men and women alternate?
- There are six different candidates for governor of a state. In how many different orders can the names of the candidates be printed on a ballot?
- In how many different orders can five runners finish a race if no ties are allowed?
- Find the number of 5-permutations of a set with nine elements
- Find the value of each of these quantities
- How many permutations of {a, b, c, d, e, f, g} end with a?
