Question 51

How many five digit 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?

Solution

Possible numbers with unit's place as 5 = $$4 \times 3 \times 2 \times 1 = 24$$

Possible numbers with unit's place as 4 and ten's place 3,2,1 = $$3 \times 3 \times 2 \times 1 = 18$$

Possible numbers with unit's place as 3 and ten's place 2,1 = $$2 \times 3 \times 2 \times 1 = 12$$

Possible numbers with unit's place as 3 and ten's place 1 = $$1 \times 3 \times 2 \times 1 = 6$$

Total possible values = 24+18+12+6 = 60

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