Number of such injective function = Total injective function - [which Pth element of set A match to Pth elemet of B]
Total injective function = 5p3= 5*4*3= 60
Now, which Pth element of set A match to Pth elemet of B
Apply inclusion exclusion formula = (n(1)+n(2)+n(3) -n(1,2)-n(1,3) - n(2,3) + n(1,2,3)) = 3*n(1) -3*(1,2) +1*n(1,2,3)
When all 3 map to pth element; Only 1 function
When any Two map to Pth element : 3c2* 3p1 = 9 function
When only 1 map to Pth element : 3c1 * 4p2 = 36 function
So,Number of such injective function = Total injective function - [which Pth element of set A match to Pth elemet of B]
= 60 - [36 -9+1]
= 32