# Thread: 3-lists taken from [3]

1. ## 3-lists taken from [3]

Consider the 3-lists taken from [3]. How many are there in which each element of [3] appears at least once?

I get $_{3}P_{3}$.

Answer the same question, but for 4-lists and 5-lists taken from [3].

For the 4-list I get $_{3}P_{3}$*3.

For the 5-list I get $_{3}P_{3}$* $3^2$.

2. Originally Posted by oldguynewstudent
Consider the 3-lists taken from [3], 4-lists and 5-lists taken from [3].?
Please define those terms. I have never seen them before.

3. Originally Posted by Plato
Please define those terms. I have never seen them before.
[3] is the set of all positive integers less than or equal to 3.

The 3-lists would then be 123 132 213 231 312 and 321.

A list is an ordered sequence of objects, and a k-list is a list of length k.
Lists are also called words. Can also be written (1,2,3) etc. Without the constraint that the elements can only appear once, (1,1,1) would also be a 3-list of [3].

4. Originally Posted by oldguynewstudent
[3] The 3-lists would then be 123 132 213 231 312 and 321.
A list is an ordered sequence of objects, and a k-list is a list of length k. Lists are also called words. Can also be written (1,2,3) etc. Without the constraint that the elements can only appear once, (1,1,1) would also be a 3-list of [3].
From that it seems to me that there are $3^3$ 3-lists from [3].
And $3^4$ in a 4-list from [3].

5. Originally Posted by Plato
From that it seems to me that there are $3^3$ 3-lists from [3].
And $3^4$ in a 4-list from [3].
Yes, that would be true except for the constraint that every element must appear at least once. ( I slightly mistated this constraint in my prior post.)

Since every element must appear at least once, for the 3-list the lists would be (1,2,3) (1,3,2) (2,1,3) (2,3,1) (3,1,2) and (3,2,1); without the constraint we could add (1,1,1) (1,1,2) (1,1,3) etc.

6. Originally Posted by oldguynewstudent
Yes, that would be true except for the constraint that every element must appear at least once. ( I slightly mistated this constraint in my prior post.)
Well then, I think in the case of 4-list from [3] we want to count the number of surjections from a set of four to a set of three. There are 36 of those.
Now with this understanding <3,2,1,1> is different from <1,2,3,1>.

If it means that those are not different then the answer in $\binom{1+3-1}{1}=3$.