Q1 PERMUTATION AND COMBINATION

How many numbers can be formed from 1, 2, 3, 4, 5, without repetition, when the digit at the unit's place must be greater than that in the ten's place?

Case 1 : units place= 1 (0 possibilities)

Case 2 : units place = 2 (tens place=1 possibility (2) and 3! Arrangements for the remaining places) = 1×3!

Case 3 : units place = 3 (tens place=2 possibilities (1,2) and 3! Arrangements for the remaining places) = 2×3!

Case 4 : units place = 4 (tens place=3 possibilities (1,2,3) and 3! Arrangements for the remaining places) = 3×3!

Case 5 : units place = 5 (tens place=4 possibilities (1,2,3,4) and 3! Arrangements for the remaining places) = 4×3!

Total = 60

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