How many ways are there to put 6 boxes in 3 crates if the boxes are distinguishable but the crates are not?
+6251 How many ways are there to put 6 boxes in 3 crates if the boxes are distinguishable but the crates are not?
You can think of this as a 3 digit number in base 6. Each digit represents how many boxes in that crate.
There are
\(6^3=216\)
combinations, however they aren't all distinguishable. As there are 3 crates any given distribution of the boxes into crates is indistinguishable within a group of 3! other distributions.
For example 312 is indistinguishable from
123, 132, 213, 231, 321
So the final number is
\(\dfrac {216}{3!} = \dfrac{216}{6} = 36\)
+129845 Here's my take on this one.......
Call the identifiable boxes
A B C D E F
And I'm assuming that every crate must contain at least one box.
We could select any 4 of the 6 boxes to put in one crate, then select 1 of the 2 remaining boxes to go into a second crate leaving the last box, by default, for the remaining crate....the number of ways to do this is :
4 1 1 6C4 * 2C1 * 1C1 = 30 ways
Or, we could select any 3 of the 6 boxes to put in one crate, then select 2 of the 3 remaining boxes to go into a second crate leaving the last box, by default, for the remaining crate....the number of ways to do this is :
3 2 1 6C3 * 3C2 *1C1 = 60 ways
Or, we could select any 2 of the 6 boxes to put in one crate, then select 2 of the 4 remaining boxes to go into a second crate leaving the last 2 boxes, by default, for the remaining crate....the number of ways to do this is :
2 2 2 6C2 * 4C2 * 2C2 = 90 ways
So....the total number of ways = 30 + 60 + 90 = 180 ways
