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I understand how to calculate permutations so:

if I have 5 people: A,B,C,D,E how many ways can you sit them down if you have 3 chairs? Answer: 5*4*3.

Also, I get that, for example, in my first chair I have 5 possible people that I can sit down, then I will have one less, so it will be 4, and lastly I have only 3 people remaining, so it is 5,4,3. What I don't understand is why do we multiply these numbers? what does the multiplication mean in this context? Why are we multiplying and not adding or subtracting, taking square root or applying log (basically anything else)

They way I think about multiplication normally is either the number of times I am doing addition using the same number (4+4+4+4+4 = 5*4) or when I want to extract a percentage of a from a number (example:the 20% of the 80% of 10 = 10 *0.8 *0.2 = 10*(4/5)*(1/5) ) but in this context I am not entirely sure what it means.

I hope I am making sense.

(I know that there are similar questions asked out there, but I am not satisfied with their answer that is why I have posted it)

Felipe Araya
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  • Have you ever heard of the rule of product? Do you understand why if I have five answers to a first question, and four answers to a second question, there are $5\times 4$ ways to answer both questions in sequence? – JMoravitz Apr 30 '20 at 21:49
  • Because for each of the $5$ ways you can pick the first person, you will have $4$ ways of picking the second. If you have $5$ things and they each have $4$ substhings, how many subthings do you have. – fleablood Apr 30 '20 at 21:49
  • Perhaps for the two dimensional case, a grid might be helpful to build intuition. If my answers to the first question are A,B,C,D,E and my answers to my second question are p,q,r,s., then the possible results are: $\begin{array}{cccc}Ap&Aq&Ar&As\Bp&Bq&Br&Bs\Cp&\dots&\dots\ \vdots &&&\vdots\ Ep&Eq&Er&Es\end{array}$. Now... you should have seen multiplication used for area of a rectangle I'd hope – JMoravitz Apr 30 '20 at 21:52
  • Check out my answer to this – Shooter Apr 30 '20 at 21:57

1 Answers1

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You have $5$ chooses for the first person. Amy, Barbara, Charlotte, Doris, and Emily.

If you choose Emily first you have $4$ ways of choosing the first two people. Emily and Amy, Emily and Barbara, Emily and Charlot, or Emily and Doris.

If you choose Doris first you also have $4$ ways of choosing the first two people.Doris and Amy, Doris and Barbara, Doris and Charlot, or Doris and Emily.

In fact, no matter whom you pick first, you will have $4$ ways of choosing the first two people. THe first person and any of the four who aren't first.

So $5$ ways of picking the first person, and each of those have $4$ ways of choosing the second, that is a total of $\underbrace{4+4+4+4+4}_{5\text{ times}}= 5*4$ ways to pick the first two people.

Consider, below are the five ways to choose the first person, and the $4$ ways to choose the second, and therefore the $20$ ways

$\begin{cases}\text{Choose A first:}&\underbrace{AB \ \ AC\ \ AD \ \ AE}_{4\text{ choices}}& 4\text{ways}\\\text{Choose B first:}&\underbrace{BA \ \ BC\ \ BD \ \ BE}_{4\text{ choices}}& 4\text{ways}\\\text{Choose C first:}&\underbrace{CA \ \ CB\ \ CD \ \ CE}_{4\text{ choices}}& 4\text{ways}\\\text{Choose D first:}&\underbrace{DA \ \ DB\ \ DC \ \ DE}_{4\text{ choices}}& 4\text{ways}\\\text{Choose E first:}&\underbrace{EA \ \ EB\ \ EC \ \ ED}_{4\text{ choices}}& 4\text{ways}\\\text{Total:}& 4+4+4+4+4& 5\times 4= 20\text{ ways}\end{cases}$

fleablood
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