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?	Pick one:
A) 54	B) 60	C) 50	D) 2 x 4!	E) 51 / 3
The digit in the unit’s place should be greater than that in the ten’s place. Hence if digit 5 occupies the unit place then remaining four digits need not to follow any order. Hence the required numbers = 4!. However, if digit 4 occupies the unit place then 5 cannot occupy the ten’s position. Hence digit at the ten’s place will be one among 1, 2 or 3. This can happen in 3 ways. The remaining 3 digits can be filled in the remaining three places in 3! ways. Hence, in all we have (3 x 3!) numbers ending in 4. Similarly, if we have 3 in the unit’s place, the ten’s place can be either 1 or 2. This can happen in 2 ways. The remaining 3 digits can be arranged in the remaining 3 places in 3! ways. Hence, we will have (2 x 3!) numbers ending in 3. Similarly, we can see that there will be 3! numbers ending in 2 and no number ending with 1.   Hence, the total number of numbers = 4! + (3 x 3!) + (2 x 3!) + 3!   = 4! + 6 x 3! = 24 + (6 x 6) = 60   ANSWER:60
<result>B</result>
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?