There are 90 people in the crowd and 10 digital music players are to be given away. As there are no restrictions on the number of music players that can be given to anyone, it is assumed that any number of music players can be given to each of the 90 people.
The number of ways of distributing each player is 90. The total number of ways of distributing the 10 players is 90^10.
In a crowd of 90 people 10 music players can be given away in 90^10 ways.
Are we assuming that the music players are different, and that matters, or is all that matters how many the people get and not which ones? The above answer assumes that we want to distinguish which player is given to whom.
As a concrete example, if we have 2 music players to distribute to 2 people, then if we don't distinguish players, there are 3 ways- person 1 gets both, person 2 gets both, or each gets one. If we distinguish music players, then there are indeed `2^2=4` ways to give them away (person 1 gets both, person 2 gets both, person 1 gets player 1 and person 2 gets player 2, person 1 gets player 2 and person 2 gets player 1).
If we don't distinguish between different music players, the problem is a little harder to solve, but there is a clever way called the "stars and bars" method.
To represent the case of both people getting a player, we use the diagram *|*, where * represent a music player and | separates the two people.
Person 1 getting both players would be represented by **| and person 2 getting both players by |**
These 3 diagrams represent the number of ways to insert a | among two ** to give away the players, in other words, 3 choose 1 ways. The 3 comes from the 3 symbols, and the 1 comes from the fact that there is one |. Note that there are 2 *'s, so another way to think of it is as 3 choose 2, which is still 3.
Now if we have 90 people, we need 89 | to separate them all, and 10 *'s to represent the music players, for a total of 99 symbols. If we have 99 slots to fill with 89 | and 10 * this can be done in
`_99 C_(10)` ways, or equivalently `_99 C_89` if you prefer.