These are the possible combinations:
This is the first group of combinations: there are 11.
21111111111
12111111111
11211111111
11121111111
11112111111
11111211111
11111121111
11111112111
11111111211
11111111121
11111111112
2211111111
2121111111
2112111111
2111211111
2111121111
2111112111
2111111211
2111111121
2111111112
If you see these are the possible solutions if there is 2 as the first one. So if you multiply the number of possible solutions with the number of possible places where the first 2 could be. The number of possible solutions is 9 multiplied by the number of places the first 2 could be is 9 because one of the places would be taken by the other 2.
9x9=81
Now if it has three 2s in it. I will give you one example:
222111111
221211111
221121111
221112111
221111211
221111121
221111112
So the possible number of places the third 2 is 7. The possible number of places places of the second 2 is 7 and, obviously, there are 7 possible places that the first 2 could be in.
7x7x7=343
Do you notice a pattern?
In the fourth one there are four 2s and four 1s. This leaves totally eight places where a 2 or a 1 could be placed. For each 2 there will be five possible places as the three other 2s would have taken up the other places.
5x5x5x5=625
If there were five 2s,then there would be two 1s, this means there are 7 possible places where you could put a 2 or a 1. For each 2 there are two possible places it could be put. This is because the other four 2s would have taken up the other places.
2x2x2x2x2=32
Now it is almost impossible to not notice the pattern.
So if there were six 2s there would be no 1s so that means there is only 1 possible solution here.
So 11+81+343+625+32+1=1093
The solution to this problem is 1093