- #1

RM86Z

- 23

- 6

- Homework Statement
- If A is a finite set, its cardinality, o(A), is the number of elements in A. Compute

(a) o(A) when A is the set consisting of all five-digit integers, each digit of which is 1, 2, or 3.

(b) o(B), where B = {x ∈ A : each of 1,2 and 3 is among the digits of x} and A is the set in part (a).

- Relevant Equations
- 3^5, 5P3, 3^2

QUESTION:

If A is a finite set, its cardinality, o(A), is the number of elements in A. Compute

(a) o(A) when A is the set consisting of all five-digit integers, each digit of which is 1, 2, or 3.

(b) o(B), where B = {x ∈ A : each of 1,2 and 3 is among the digits of x} and A is the set in part (a).

SOLUTIONS:

a) 3^5 = 243 (correct answer as per the book).

b) There is no answer in the book and I am not entirely sure if I am on the right track as below:

As I see this there are five places for the digits 1,2 and 3 to go so we take the permutation of the number of places taken three times; 5P3 = 5!/2! = 60. Then there are two places left over, each of which has 3 choices from {1,2,3} for a total of 3^2 = 9. So the result is 5P3 x 3^2 = 60 * 9 = 540. But this is a larger than o(A) so this cannot be correct?

If A is a finite set, its cardinality, o(A), is the number of elements in A. Compute

(a) o(A) when A is the set consisting of all five-digit integers, each digit of which is 1, 2, or 3.

(b) o(B), where B = {x ∈ A : each of 1,2 and 3 is among the digits of x} and A is the set in part (a).

SOLUTIONS:

a) 3^5 = 243 (correct answer as per the book).

b) There is no answer in the book and I am not entirely sure if I am on the right track as below:

As I see this there are five places for the digits 1,2 and 3 to go so we take the permutation of the number of places taken three times; 5P3 = 5!/2! = 60. Then there are two places left over, each of which has 3 choices from {1,2,3} for a total of 3^2 = 9. So the result is 5P3 x 3^2 = 60 * 9 = 540. But this is a larger than o(A) so this cannot be correct?