I am trying to find a relation to the following two problems:
Problem 1: Mary bought $14$ different baseball cards, and wants to give all of them to her son Jack over $7$ days. In how many ways can she do this?
$\underline{Solution}$:
The total number of ways she can do this $7^{14}$ since there are $n^k$ ways to distribute $k$ distinguishable objects into $n$ distinguishable boxes. QED
Problem 2: How many $4$-letter words can we make with $26$ letters?
$\underline{Solution}$:
There are $26^4$ ways of making $4$-letter words with $26$ letters, as for the first slot of the $4$-letter word, there are $26$ choices of letters, for the second slot of the $4$-letter word, there are $26$ choices of letters, etc. So we get $26\times 26 \times 26 \times 26=26^4$ by the multiplication rule. QED
Now I would like to relate the two problems since both solutions are of the same form. Therefore for:
Problem 1 I can say the first card has $7$ choices of days to be given on, the second card has $7$ choices of days to be given on, etc. So there are $7 \times 7 \ldots \times 7=7^{14}$ ways of distributing $14$ cards over $7$ days.
Problem 2 I can say there are $4$ distinguishable objects (positions of each letter in the $4$-letter i.e. slot$_1$, slot$_2$, slot$_3$, slot$_4$) and $26$ distinguishable boxes (letters) so there are $26^4$ ways of distributing $4$ slots into $26$ letters.
In Problem 1 it makes sense to think of it both ways, but for Problem 2 we clearly have to do some mental acrobatics for it to make sense, which could lead to some confusion about what to consider boxes and what to consider objects. Namely, I am confused about how to think of the following problem:
A retired person chooses randomly one of the six distinguishable parks of his town every day and spends the day there hiking. How many different ways can he spend $10$ days?
I want to say $6^{10}$ as the first day he has $6$ choices, on the second day he has $6$ choices, etc. But I also want to say $10^6$ because this smells like the Problem 1. PLEASE HELP ME
Anything you offer is greatly appreicated. I am losing my mind
In "Problem 1", the number of choices is $7$ while the number of times the choice is made is $14$. In "Problem 2", there are $26$ choices of letter and you make that choice $4$ times, so there are $26^4$ possible 4-letter outcomes. For the retired person, the number of choices is $6$ while the number of times he must make that choice is $10$. So, you are correct that it is $6^{10}$.