Answer:
Consider a set {1,2,3}
Since f is onto, all elements of a set will have unique pre-images.
Element |
Number of possible pairings |
1 |
3 |
2 |
2 |
3 |
1 |
Total number of one-one function = 3 × 2 × 1 = 6
Now consider a set {1,2,3………n}
Since f is onto, all elements of a set will have unique pre-images.
Element |
Number of possible pairings |
1 |
3 |
2 |
2 |
3 |
1 |
. |
. |
. |
. |
n – 1 |
2 |
n |
1 |
Total number of one-one function = n × (n – 1) × (n – 2) × …….× 2 × 1 = n!