I have 32 slots and 8 balls.
A slot can have minimum 0 balls and maximum 1 ball.
All the 8 balls can be present in any slots, or it can be less than 8. It is also possible that no balls are present in any slot.
For example -
Slot 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
Cmbn 1 B 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Cmbn 2 B B B B B B B B 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Cmbn 3 0 B B B B B B B B 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Cmbn 3 0 0 B B B B B B B B 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Cmbn 4 B 0 B B B B B B B 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Cmbn 5 B 0 0 B B B B B B 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Cmbn 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 B B 0 0 B B
Cmbn 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 B B 0 B B 0 0 B B
Cmbn 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 B B B B B B B B 0 0 0 0 0 0 0
Cmbn 9 B 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 B B B B B B B B 0 0 0 0 0 0 0
Cmbn 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
...
How many such combinations are possible?
What is the formula applicable for this scenario?
Thanks.
$$\sum_{i=0}^8 \dbinom{32}{i} = 15,033,173$$
What this formula means: $\dbinom{32}{0} = 1$ is the number of ways to choose zero slots to have balls, $\dbinom{32}{1} = 32$ is the number of ways to choose exactly $1$ slot to have a ball, $\dbinom{32}{2} = 496$ is the number of ways to choose two of the $32$ slots to have balls...
In general: $$\dbinom{n}{k} = {_n}C_k = \dfrac{n!}{(n-k)!k!}$$
Example: $\dbinom{32}{8} = \dfrac{32!}{24!8!} = \dfrac{32\cdot 31\cdot 30\cdot 29\cdot 28\cdot 27\cdot 26\cdot 25}{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1} = 10,518,300$