Question

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?

• A. 54
• B. 60
• C. 17
• D. \( 2 \times 4! \)

Hint:

For problems with number arrangements and constraints like this, first calculate the number of valid pairs and then consider the remaining positions for the other digits.

Answer 1

The Correct Option is B

We need to form a two-digit number such that the digit in the unit's place is greater than the digit in the ten's place. For the ten’s place, we can select any digit from 1 to 4 (since the unit's place must be greater). For each choice in the ten's place, there are 4 remaining digits available for the unit’s place, but only those greater than the chosen digit in the ten's place. - If the ten’s digit is 1, the unit's digit can be 2, 3, 4 (3 choices). - If the ten’s digit is 2, the unit's digit can be 3, 4 (2 choices). - If the ten’s digit is 3, the unit's digit can be 4 (1 choice). Thus, the total number of such pairs is: \[ 3 + 2 + 1 = 6. \] Now, we can form a 2-digit number for each pair, so we multiply by the number of ways to arrange the remaining digits: \[ 6 \times 4! = 6 \times 24 = 60. \]