If we make a sequence selecting three elements from three different elements {1,2,3} and we permit overlapped elements for the sequence, then the total number of sequences is [?]. If we do not take into account the order, the total number of the selections is [?]

Question

triciaal · Answer

@mathstudent55 ?

holsteremission · Answer

What does "overlapped" mean here? Like the sequence $\{1,2,1\}$ is allowed? (i.e. elements from the set are taken and replaced)

styxer · Answer

@HolsterEmission I'm more confused about the overall statement than the math itself.

Since the answer for the first part is "27", I assume that we can use any number (1,2,3) and they can repeat, so 3x3x3 = 27

As for the second part of the question, the answer is 10... I just don't understand what the question is itself "/

holsteremission · Answer

Okay, that's reasonable enough. The word choice is questionable.

The reasoning for the first part is correct. Any position in a sequence of three elements has three possible candidates for that position because each position can be chosen independently of the others.

These sequences are
$$\begin{array}{ccc}
\{1,1,1\}&\{1,1,2\}&\{1,1,3\}\
\{1,2,1\}&\{1,2,2\}&\{1,2,3\}\
\{1,3,1\}&\{1,3,2\}&\{1,3,3\}\
\hline
\{2,1,1\}&\{2,1,2\}&\{2,1,3\}\
\{2,2,1\}&\{2,2,2\}&\{2,2,3\}\
\{2,3,1\}&\{2,3,2\}&\{2,3,3\}\
\hline
\{3,1,1\}&\{3,1,2\}&\{3,1,3\}\
\{3,2,1\}&\{3,2,2\}&\{3,2,3\}\
\{3,3,1\}&\{3,3,2\}&\{3,3,3\}\end{array}$$Put another way, this is the same as the number of three-digit numbers you can make with the digits $1,2,3$.

For the second part, "not taking order into account" means we omit any of the possibilities above that use the same digits. By this I mean the sequences $\{1,1,2\}$, $\{1,2,1\}$, and $\{2,1,1\}$ are considered the same and together count as one.

styxer · Answer

@HolsterEmission So, I assume that (1,1,1) ,  (2,2,2) and (3,3,3) are unique selections.
And you stated that using the same digits counts as same, so we have:
6 selections that are just one (1,2,3) (1,3,2) (2,1,3) (2,3,1) (3,1,2) (3,2,1) 
and the other (27-3-6) 18 sequences will have two equal numbers and one different number, switching the position 3 times, so 18 sequences are in fact just 6.
6+3+1 = 10 sequences. Is that correct?

The statement is really confused for me, how I can interpretate this question?

holsteremission · Answer

That's correct. The second question basically comes down to being able to differentiate between permutations and combinations. The telling difference between them is that permutations account for order, while combinations do not. So generally, the number of permutations is greater than the number of combinations. You can learn more about this with several resources. Here's the top result from a search for "combinations vs. permutations": http://www.regentsprep.org/regents/math/algtrig/ats5/lcomb.htm

holsteremission · Answer

In your question, you're looking at the possible combinations.

styxer · Answer

@HolsterEmission I see. I thought that "sequence" would mean a specific order like 1,2,3 or 3,2,1... But it's any possible arrangement of numbers following the rules...

The "overlapped" is for repeated elements being different sequences , and the "don't take into account the order" means that (1,2,3) = (3,2,1) , just like you said. 

Thank you very much!