CUET Preparation Today
Set A has 3 elements and the set B has 4 elements. Then the number of injective mappings that can be defined from A to B is |
12 24 64 144 |
24 |
The correct answer is Option (2) - 24 No. of injective mappings = select any 3 from set B × No. of ways to arrange them in front of A $={^4C}_3×3!$ $=24$ |
