There are 6 maps f from {1,2,3,4,5,6} to itself such that the image of f contains 1, 2, and 3.
To count the number of maps f from the set {1,2,3,4,5,6} to itself such that the image of f contains 1, 2, and 3, we can consider the possibilities for the elements 4, 5, and 6 in the image.
Since we want the image of f to contain 1, 2, and 3, these elements are fixed. Now, we have three remaining elements (4, 5, and 6) that can be mapped to any of the remaining three elements in the range.
For the element 4, there are 3 choices left (it can be mapped to any of the three remaining elements in the range).
For the element 5, there are 2 choices left.
For the element 6, there is only 1 choice left.
The total number of such maps is the product of the choices for each element:
3×2×1=6
So, there are 6 maps f from {1,2,3,4,5,6} to itself such that the image of f contains 1, 2, and 3.