Question

Suppose you have 9 different (distinguishable) coins, and 4 different (distinguishable) jukebox
slots.
(i) How many ways can you insert the 9 coins into the jukebox slots, if the order in which
the coins are inserted does not matter?
(ii) How many ways can you insert the 9 coins into the jukebox slots, if the order in which
the coins are inserted into each jukebox does matter?
(iii) How many ways can you insert 6 of the coins into one of the jukebox slots, if the order
in which the coins are inserted matters?

Answer 1

(i) We can make an equivalent formulation of the problem: Put 9 numbers into an arbitrary order (9! ways to do that due to the multiplication rule) and then split them into 4 parts (some parts can be empty, but all 4 parts must contain 9 numbers). The number of possible splittings was calculated manually. It is equal to $((9\cdot8+8\cdot7+7\cdot6+6\cdot5+5\cdot4+4\cdot3+3\cdot2+2)=220$ . The total number of ways to split 9 coins between 4 slots including the order is: $220\cdot9!$

(ii) For each coin there are 4 different jukebox slots. We use the multiplication principle (https://www3.nd.edu/~apilking/Math10120/Lectures/Solutions/Topic03.pdf) and get $n=4^9$ different ways to put coins into slots.

(iii) We assume that there is a fixed jukebox slot, in which we are going to put 6 coins. Using the multiplication principle, we get $6!=720$ different ways.