There are at most 5 1s (pigeon hole principle)

(1) Put 1 in slot 1 and put another 1 in slot 3, 4,...,10 repeatedly, we get 8 numbers.

Next put 1 in stot 2,  and put anothe 1 in slot 4,...,10 repeatedly, we get 7 numbers.  In this way, we will get 8+7 +6 ... +1=36.

(2)Now consider 3 1s.  Put 1 in slot 1 and put anothe 1 in slot 3, put the 3rd 1 in slot 5, 6,...,10, we get 6 numbers.  Then move the 1 from slot 3 to slot 4.  put the 3rd 1 in slot 6,...,10, we get 5 numbers. In this way, we will get 6+... +1=21.

(3)Next Put the first 1 in stot 2, and put the second 1 in slot 4.  Repeat step (2), we will get another 21 numbers.

(4) Now keep the first 1 in stot 2, put the second 1 in slot 5 and put the 3rd 1 in slot 7,...,10., we get 4 numbers.  Move the second and the third 1s as in step (2) , we will get additional 3,2 and1 numbers.  We get 4+...+1=10.

The sum now is 36+2x21+10=88.  The 5 1s will yield another number, bring the total to 89.  The 3 1s case is just a partial result since if we put the first 1 in slot 3, we will get more numbers.  When we use 4 1s, we will get even more numbers.