Discrete Math - Permutation, Combination

Question

So for this question, would it be 2C1 or 2P1 multiplied by 6P3 or 6C3 and why? It's not clear to me if I am using permutation or combination for this question. Thank you for your help!

Answer 1

Giving more detail to Alex's answer: suppose our six people are named Alfred, Bobby, Clarissa, Devon, Erin, and Francis (shortened hereafter to just their first initial: a,b,c,d,e,f)

They will stand in the following pattern:

$$\begin{array}{|c|c|c|}\hline 1&2&3\\ \hline 4&5&6\\ \hline\end{array}$$

Here, we consider two arrangements of people to be different if any location has a different person standing there between these two arrangements.

For example: $\begin{array}{|c|c|c|}\hline A&*&*\\\hline *&*&*\\\hline\end{array}$ is considered a different arrangement of people than $\begin{array}{|c|c|c|}\hline B&*&*\\\hline *&*&*\\\hline\end{array}$ regardless of how the rest of the spaces are filled because the person standing in the top left is different in the one than in the other.

We count how many arrangements there are via multiplication principle.

• Pick who stands in position $6$. There are six choices
• Pick who stands in position $5$. As we cannot pick whoever we picked last step, there are $5$ remaining choices.
• Pick who stands in position $4$. As we cannot pick whoever we picked either of the last two times, there are $4$ remaining choices.
• $\vdots$
• Pick who stands in position $1$. There is only one choice left available.

There are then $6\cdot 5\cdot 4\cdots 1 = 6!$ possible arrangements.

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This is a different question than the one where we don't care about position within a row were we could consider $\begin{array}{|c|c|c|}\hline A&B&C\\\hline D&E&F\\\hline\end{array}$ to be the "same arrangement" as $\begin{array}{|c|c|c|}\hline B&C&A\\\hline F&D&E\\\hline\end{array}$.

If order within each row doesn't matter, then simply pick who the three people are who stand in the front row. There are $\binom{6}{3}$ ways to accomplish that. As there is no other relevant information which differentiates between different arrangements, that would be the total.

This is again a different question from the one where we consider only the front row of people and how they are arranged, having the back row invisible and unable to be seen by the camera. In that case, pick who stands front left, pick who stands front middle and pick who stands front right. As no other information will differentiate further, we have the total number of arrangements as being $6\cdot 5\cdot 4$.

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Long story short, interpretation of a problem is incredibly important in combinatorics and it should be made perfectly unambiguous as to what is being asked and what is meant by "different arrangements." The most natural definition of "different arrangements" would be the one given above.

Furthermore, don't worry so much about "when do I use permutations" and "when do I use combinations" as every question can be broken down into more basic components. Learn multiplication principle and how to use it. All permutation or combination questions can have their answer be found from first principles and multiplication principle without having to even touch permutations or combinations.

Answer 2

Presumably order matters within each row, and also row matters. In this case the answer would be $6!$.