In how many ways can some or all of the $5$ distinct coins be put into
$8$ pockets?
Could this be modeled as the problem of "In how many ways N distinct items be put into r distint groups (where some groups may be empty)?"
My instructor is of the opinion that this problem should be modeled as like for every coin we have $9$ options to put it into one of the 8 pockets or don't and so the answer is $9^5$,however in this case I think $-1$ is necessary to avoid the case when none is selected but he thinks we don't need that.
There is also a different model that I think could be the solution which is as a summation of $8^1 + 8^2 + \cdots + 8^5$ pertaining to how many ways $1$ coin could be distributed in $8$ pockets then $2$ coins in $8$ pockets … till $5$ coins in $8$ pockets.(assuming pockets are distinct in every case)
But as the answer is not given I can't be sure which is correct,(instructor haven't disclosed it either),could any body tell me which one is correct,if none is correct,then how exactly I am supposed to count this?
Best Answer
If I understand correctly, for each of $\binom{5}{k}$ choices of $k$ coins, you will have $8^k$ maps from those $k$ coins to the $8$ pockets. If that is the case, then the answer would be $$ \sum_{k=0}^5\binom{5}{k}8^k = (1+8)^5 $$ by the binomial theorem.
Another way of looking at this, is to add a ninth hidden pocket for the coins that don't appear in the visible $8$ pockets. That is, you are asking how many ways to put $5$ coins into $9$ pockets, or $9^5$.
If you don't want to allow $8$ empty (visible) pockets, then the answer would be $9^5-1$.