No of n digits no in incresing order i have a set{1,2,3}, now i want to find no of n digits no such that they always start with 1 and ends with 3 in  non decreasing order and all nos from set must be included atleast by 1.
for ex. if n=4 then, 1123,1223,1233 are only combination, so ans is 3.

Now, in general terms, consider a set(1,2,3....,k} ,then what should be the answer such that they starts with 1 and ends with k and they must be in non decreasing order.
The numbers in the $n$ digits must be ordered as $1,2,\cdots,k$. Let $a_t$ be the number of $t$s in the $n$ digits. Then $a_1+a_2+\cdots+a_k=n$ and $a_t\geq1$ for all $1\leq t\leq k$. And for each such combination of $a_t$s, there corresponds exactly one $n$ digit number satisfying the condition.(112222333 corresponds to 2+4+3=9)

So the number of possible $n$ digit numbers are same as the number of possible $a_t$s. Now $(a_1-1)+(a_2-1)+\cdots+(a_k-1)=n-k$ and each term in the summation should be a nonnegative integer. Each solution of this equation corresponds to one arrangement of $(k-1)$ '$+$'s and $(n-k)$ '$1$'s and vice versa.(For example, 1+3+2=6 corresponds to 1+111+11) Thus the number of solutions to the equation is $\binom{n-1}{k-1}$.
As karvens pointed out, my original answer on this was wrong. It is, in fact, a "stars and bars"  situation, as per his solution. We have "bars" representing the divider "|" between each type of digit and "stars" representing each digit. E.g. the string "$|*|**|**|$" represents the number "12233".

We must have a bar at the end of the string and each of the $k$ digits must appear once so we are arranging $k-1$ lots of "$|*$" and $n-k$ lots of "$*$". So we have $k-1 + n-k = n-1$ objects in total and we need to choose places for the $k-1$ lots of "$|*$".

The number of ways to do this is $$\binom{n-1}{k-1}$$