# 3-lists taken from [3]

• May 18th 2010, 03:25 AM
oldguynewstudent
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 \$\displaystyle _{3}P_{3}\$.

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

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

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

• May 18th 2010, 06:27 AM
Plato
Quote:

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.
• May 18th 2010, 07:39 AM
oldguynewstudent
Quote:

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].
• May 18th 2010, 07:52 AM
Plato
Quote:

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 \$\displaystyle 3^3\$ 3-lists from [3].
And \$\displaystyle 3^4\$ in a 4-list from [3].
• May 18th 2010, 08:05 AM
oldguynewstudent
Quote:

Originally Posted by Plato
From that it seems to me that there are \$\displaystyle 3^3\$ 3-lists from [3].
And \$\displaystyle 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.
• May 18th 2010, 08:20 AM
Plato
Quote:

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 \$\displaystyle \binom{1+3-1}{1}=3\$.