Determine remainder when N is divided by 1000?

Question

Question:

If N denotes the number of $7$ digit positive integers having property that their digits are in non-decreasing order then, find $N \pmod {1000}$

(assume that repeated digits are allowed)

My attempt:

I am familiar with the case when repetition is not allowed in that case answer is simply found by picking 7 numbers out of 9 in $^9C_7$ =36 ways (or by removing 2 digits from any 9 digit number ,having digits sorted previously in increasing order,in $^9C_2$ ways).

but, I don't know how to do the aforesaid question in which repetition of digits is allowed with only restriction that digits from left to right must be in ascending order.

edited: replaced "increasing order" with "non-decreasing order".

Answer 1

This is an answer if repetition of digits is allowed and the digits are demanded to be in non-decreasing order (weaker than increasing order).

For $i=1,\dots,9$ let $n_i$ denotes the number of occurences of digit $i$.

Then $N$ equals the number of sums $n_1+\dots+n_9=7$ where the $n_i$ are nonnegative integers.

E.g. solution $\langle 1,0,3,0,0,2,0,0,1\rangle$ corresponds with number $1333669$.

This can be solved with stars and bars and we find:$$N=\binom{7+8}{8}$$

Answer 2

Hint : If digits of the number are in increasing order, none of the two digits can be same. So the problem is equivalent to find number of such combinations, with no repetition of digits at all.