Combination problem technique 
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. 
 A: 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$
$6 + 12 + 18 + 24 = 60$
A: 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:


*

*The number of combinations that end with $12$ is equal to $3!=6$

*The number of combinations that end with $13$ is equal to $3!=6$

*The number of combinations that end with $14$ is equal to $3!=6$

*The number of combinations that end with $15$ is equal to $3!=6$

*The number of combinations that end with $23$ is equal to $3!=6$

*The number of combinations that end with $24$ is equal to $3!=6$

*The number of combinations that end with $25$ is equal to $3!=6$

*The number of combinations that end with $34$ is equal to $3!=6$

*The number of combinations that end with $35$ is equal to $3!=6$

*The number of combinations that end with $45$ is equal to $3!=6$


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:


*

*The number of combinations that end with $21$ is equal to $3!=6$

*The number of combinations that end with $31$ is equal to $3!=6$

*The number of combinations that end with $32$ is equal to $3!=6$

*The number of combinations that end with $41$ is equal to $3!=6$

*The number of combinations that end with $42$ is equal to $3!=6$

*The number of combinations that end with $43$ is equal to $3!=6$

*The number of combinations that end with $51$ is equal to $3!=6$

*The number of combinations that end with $52$ is equal to $3!=6$

*The number of combinations that end with $53$ is equal to $3!=6$

*The number of combinations that end with $54$ is equal to $3!=6$


So the number of legal combinations is $120-(6+6+6+6+6+6+6+6+6+6)=60$
