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Suppose you have $20$ different rings and $4$ display towers. On each tower the rings are stacked one above another. In how many ways can they be arranged if:

[a]: The order of rings on each tower does not matter:

[b]: The order of rings on each tower matters, and there are exactly 5 rings on each tower?

[c]: The order of rings on each tower matters, and each tower can hold any number of rings?

Here is what I am thinking, not sure if it's right:

[a]:

Order does not matter - standard combination question:

$$Ni = # of rings on tower i$$ then: $$N1+N2+N3+N4 = 20$$ and we find number of solutions to this problem I.E (Stars and bars): $$\binom{20+3} {3} = 1771$$

[b]: I know that this is a permutation question since order is important, but I am not sure if I am doing it right: $$ 20P5 * 4 $$ [c] have no clue

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For [a], what matters is what tower each ring is put on. The first ring can go on each of the four towers, so can the second, and so on. So the result is $4^{20}$.

For [b], you can just order all $20$ rings in a single row, then put the first five on the first stand, rings number 6 through 10 on the second stand, and so on. So the answer is $20!$

For [c], it's a stars and bars solution, only the internal order of the stars and of the bars matter. The answer then comes out to be $23!$

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