How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition), when the digit at the units place must be greater than that in the tenth place?
It can be easily solved that, the total combination is 5! .
Now if i obey the condition, i might have to fill the unit digit by 5 , 4, 3 and 2. I can not put 1 because there is no greater number than this. But how to add the the total. I messed up totally.
On
When it is a 2 in the units place the tens digit can only be a 1. There are 6 ways to distribute the other 3 numbers. $ 3\cdot 2\cdot 1 \cdot 1 \cdot 1 = 6 $
When it is a 3 in the units place the tens digit can be a 1 or 2. There are 6 ways to distribute the other 3 numbers.$ 3\cdot 2\cdot 1 \cdot 2 \cdot 1 =12$
When it is a 4 in the units place the tens digit can be a 1 or 2 or 3. There are 6 ways to distribute the other 3 numbers.$ 3\cdot 2\cdot 1 \cdot 3 \cdot 1 =18$
When it is a 5 in the units place the tens digit can be a 1 or 2 or 3 or 4. There are 6 ways to distribute the other 3 numbers.$ 3\cdot 2\cdot 1 \cdot 4 \cdot 1 =24$
$6 + 12 + 18 + 24 = 60$
Based on your statement "the total number of combinations is $5!$", I am left to assume that you meant to ask "how many $5$-digit numbers can be formed?"...
Option #$1$ - count the number of legal combinations:
So the number of legal combinations is $6+6+6+6+6+6+6+6+6+6=60$
Option #$2$ - count the number of illegal combinations and subtract it from the total:
So the number of legal combinations is $120-(6+6+6+6+6+6+6+6+6+6)=60$