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?

Answer 1

$\huge\mathfrak{Solution}$
$<b>$
According to the question, the digit in the unit’s place should be greater than that in the ten’s place.

Hence, if 5 is placed at the unit place, then remaining four places can be filled with any of the

four digits 1, 2, 3 and 4, hence total numbers =4!

However, if digit 4 is placed at the unit place then 5 cannot occupy the ten’s place.

Hence, the digits at the ten’s place can be 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 × 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 × 3!) numbers ending in 3. Similarly, we have 3! numbers ending in 2 and no number ending with 1. Hence total number of numbers=  4! + 3 × 3! + 2 × 3! + 3! = 24 + 18 + 12 + 6 = 60.

Thank you

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Answer 2

23 24 25
34 35
43 45