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?
Explanation:
tTen's place uUnit's place
u > t
Case (i): t = 1 u can be 2, 3, 4, 5 ⇒ 4 possibilities So total possible number that can be formed = 4 × 3 × 2 × 1 Case (ii): t = 2 u can be 3, 4, 5 ⇒ 3 possibilities So total possible number that can be formed = 3 × 3 × 2 × 1 Case (iii): t = 3 u can be 4, 5 ⇒ 2 possibilities So total possible number that can be formed = 2 × 3 × 2 × 1 Case (iv): t = 4 u can be 5 ⇒ 1 possibility So total possible number that can be formed = 1 × 3 × 2 × 1 Hence, total possible number = (4 + 3 + 2 + 1) 3 = 60
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