[Math] Number of combinations for heteregeneous slots

combinationscombinatorics

Definition: Suppose you have a slot machine with N-number of slots. These slots have the property of being heterogeneous. That is, within each slot, the set of possible outcomes for slot i, $s_i$ are disjoint, i.e. $s_i \cup s_j = \emptyset$. Consequently, it is possible that $|s_i| \neq |s_j|$, where $|s_i|$ is the number of elements in the set.

Example: A slot machine has 2 heterogeneous slots, where slot 1's possible results are $\{a, b, c, d\}$ while slot 2's are $\{1, 2, 3\}$.

Knowledge: When the slot machine is played, the number of possible outcomes can be calculated as $\Pi_{i=1}^{N} |s_i|$. This is the number of combinations, since order doesn't matter. Also, no repetitions are allowed. For the example above, the number of combinations for the 2 slots are $4 \times 3 = 12$. The actual possible combinations are listed below for clarity.

{a, 1}
{a, 2}
{a, 3}
{b, 1}
{b, 2}
{b, 3}
{c, 1}
{c, 2}
{c, 3}
{d, 1}
{d, 2}
{d, 3}

Special case: Suppose any element in the slot sets could be replaced with one unique homogeneous value, e.g. BLANK. Following the example above, the combinations for this special case (ignoring repetitions) would be as follows.

{BLANK, 1}
{BLANK, 2}
{BLANK, 3}
{a, BLANK}
{b, BLANK}
{c, BLANK}
{d, BLANK}
{BLANK, BLANK}

Question: How do we calculate the number of combinations (excluding repetitions) for the special case above?

Caveat: For the case where $N = 2$, the solution seems to be $|s_1| + |s_2| + 1$. However, I'm interested in a general solution for any $N \geq 1$

3-slot example: The accepted answer was cross-checked using a 3-slot scenario, where $s_1 = \{a,b,c\}$, $s_2 = \{1,2\}$, and $s_3 = \{x\}$. The manually generated 18 combinations are as follows for anyone interested in cross-checking.

{NULL, 1, x}
{NULL, 2, x}
{a, NULL, x}
{b, NULL, x}
{c, NULL, x}
{a, 1, NULL}
{a, 2, NULL}
{b, 1, NULL}
{b, 2, NULL}
{c, 1, NULL}
{c, 2, NULL}
{NULL, NULL, x}
{NULL, 1, NULL}
{NULL, 2, NULL}
{a, NULL, NULL}
{b, NULL, NULL}
{c, NULL, NULL}
{NULL, NULL, NULL}

Best Answer

You now have $s_i+1$ choices at position $i$ of the machine, but it appears you require at least one blank. The number of combinations without the requirement is $\Pi_{i=1}^{N} |s_i+1|$ by the same logic as before. The number if you require at least one blank is the number of new ones: $\Pi_{i=1}^{N} |s_i+1|-\Pi_{i=1}^{N} |s_i|$

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