Conditional combination I have 3 objects A, B and C
and 3 baskets 1, 2 and 3
A can go to any of the basket
But B can only go to the same basket with A or in the baskets that has higher number than what A chose.
C can only go to the same basket with A or B or in the baskets with higher number than what A and B chose
For example, if A is in basket 1, B can go to any of the 1,2,3. If B chose basket 2 then C can either go to 2 or 3. If A is in basket 3, B and C then have no choice and will have to go to basket 3.
I was struggling how to formulate it in a general way to find the total number of combination, so that I can do it for any number of objects and baskets.
 A: The general problem is:  "how many non-decreasing sequences of length $n$ can be formed from $\{1,\cdots, k\}$".
Call the answer $F(n,k)$
This is a Stars and Bars problem in disguise.  Indeed, every such sequence can be written (uniquely) as $$1^{a_1}2^{a_2}\cdots k^{a_k}$$
Where the $a_i$ are non-negative integers and $a_1+a_2+\cdots +a_k=n$.
(Note:  as to notation, the $a_i's$ just tell us how often each character $i$ appears.  Thus $1^32^2$ denotes the sequence $11122$, for example).
But Stars and Bars enables us to count the number of the good $k-$tuples $(a_1, \cdots, a_k)$, and we get the final answer $$F(n,k)=\binom {n+k-1}n$$
Check:  the stated problem has $n=3,k=3$ so we get $\binom 53=10$ which is easily confirmed (by just writing out all the cases, say).
A: i think your reasoning is correct you could write something like this $$ 1 1 1 ,112,113,122,123,133,222,223,233,333$$
By the way in the condition of C there is a little thing that C can go A or B or ... which adds $121,131,232$ to our possible combinations but i think it is a typo :D
So $111$ means that A,B,C all go to basket $1$ similarly if we had $321$ A goes to basket 3, B goes to 2 etc.
