How many different ways can you put 3 distinguishable things in 10 boxes?
Assuming the 10 boxes are distinguishable or ordered.
There are 10 ways to put the first item, 10 for the second, and 10 for the third. So by the multiplication rule, there are 10^3 ways to place the three objects.
Okay, but what if the maximum that each box can hold is 2?
What do you do when you have the third object to be placed and supposed to place in a box already filled with two object?
You would skip that case.
How many of these cases do we have?
To determine the number of different ways you can put 3 distinguishable things in 10 boxes, we can use the concept of permutations.
In this case, we have 3 distinguishable things (let's represent them as A, B, and C) and 10 distinguishable boxes (represented as B1, B2, B3, ..., B10).
To solve this problem, we can think of each distinguishable thing as a distinct choice for each box. So, for the first box, we have 3 choices (A, B, or C). For the second box, we have 2 choices (since we have already placed one object in the first box). Finally, for the third box, we have only 1 choice (since we have already placed two objects in the previous two boxes).
Using the fundamental principle of counting, we multiply the number of choices together:
3 choices for the first box * 2 choices for the second box * 1 choice for the third box = 3 * 2 * 1 = 6
Therefore, there are 6 different ways to put 3 distinguishable things in 10 boxes.