this can be solved bypermutation and combination based problem How many three are there whose hundred digit is greater than tens digit which in turn is greater than the unit digit?
Ans:I tried it But couldn't solve..
 A: I guess you meant how many "three digit numbers are there (...)" ?
In that case you're looking for integers that can be written as $\overline{a_2a_1a_0}^{(10)}$ such that $a_2>a_1>a_0$.
In such a situation, $a_2\in[2,9]$, $a_1\in[1,8]$ and $a_0\in[0,7]$.
Proof 1:
Whenever you choose a value for $a_2$, you have to choose $a_1$ and $a_0$ between the remaining values in the set $[0,a_2-1]$ in which there is $a_2$ integers.
Picking those two values is like picking any two different values from the remaining set and ordering them later. So the number of combinations for each $a_2$ is equal to:
$${a_2 \choose 2}=\frac{a_2(a_2-1)}{2}$$
So the total number of combinations is:
$$n=\sum_{i=2}^9\frac{i(i-1)}{2}=\frac{1}{2}\left(\sum_{i=2}^9i^2-\sum_{i=2}^9i\right)$$
We know that:
$$\cases{\sum_{i=1}^9i=\frac{9(9+1)}{2} \\ \sum_{i=1}^9i^2=\frac{9(9+1)(2\times9+1)}{6}}$$
So:
$$\cases{\sum_{i=2}^9i=\frac{9(9+1)}{2}-1=44 \\ \sum_{i=2}^9i^2=\frac{9(9+1)(2\times9+1)}{6}-1=284}$$
Now let's go back to $n$:
$$n=\frac{1}{2}\times(284-44)=120$$
Proof 2:
You could have just said that choosing three ordered integers from the $10$ possible values in $[0,9]$ is just choosing three distinct integers from that set and ordering them later. So it would be equal to:
$$n={10 \choose 3}=120$$
Proof 3:
You can also say that the number of combinations for any 3 different digits is $10\times9\times8=720$. And that if you have 3 elements you have 6 different ways of ordering them. So your total number of distinct $3$-digits numbers is:
$$n=\frac{720}{6}=120$$
Just pick whatever proof you're most comfortable with ;)
A: Hint: Could the number of ways of choosing $3$ digits be related? 
A: Hint:  If by greater you mean strictly greater (equal is not permitted) you could look for how many three digit strings there are with all digits different.  What fraction of them will have the digits in the proper order?  No leading zero is automatically assured.
