people by p1 p2 and p3 Let A 1 2 3 4 5 Consider a subset of

people by p1, p2, and p3. Let A = {1, 2, 3, 4, 5}. Consider a subset of A that contains exactly two elements. For Instance, {1, 2} is such a subset. Call such a subset a \'2-suDset\' of A. Similarly, consider a subset of A that contains exactly three elements. For instance {1, 2, 3} is such a subset. Call such a subset a \'3-subset\' of A. Let B equal the set of all 2-subsets of A, and let C equal the set of all 3-suosets of A. Explain why there is or is not a bijection between the sets B and C.

Solution

1.

3 people to be placed in 3 positions. This is done in 3! ways ie 6 ways

The 6 ways are mentioned below.

c                         s                                  w

p1                      p2                                p3

p1                      p3                                p2

p2                       p1                               p3

p2                       p3                               p1

p3                       p2                               p1

p3                       p1                               p2

2.

Number of 2 subsets are:

C(5,2)=10

Number of 3 subsets are:

C(5,3)=C(5,2)=10

So number of elements in 2 subsets and 3 subsets are equal so there is a bijection between the two sets.