Question

Is there an easy way to determine whether to use Permutations, Combinations or some differentiation thereof?

For Example,
A coach randomly selects 3 swimmers from a team of 8 to swim in a heat. What is the probability that she will pick the three strongest swimmers?

Answer 1

this is combinatorics? or whatever that thing was called

Answer 2

Theoretical and Experimental probability.

Answer 3

hmm @zepp wanna help?

Answer 4

i called you coz you rhyme with help lol

Answer 5

Oh my goodness, @lgbasallote You know that I hate probabilities :(

Answer 6

oh wow..i didnt know that..what are the odds! lol

Answer 7

Permutations are used when the order is important
combinations are used the ordering is not significant

Answer 8

what unklerhaukus said. you really have to tell from the context. and of course sometimes it is neither, or a combination of the two

Answer 9

of course i could plug this into my calculator using the nCr key but how would i solve this by hand?

Answer 10

hold the phone
i didn't read the question, but now i see you have to pick exactly the three strongest swimmers out of the 8

Answer 11

there are two ways to do this. one way is to compute how many ways she can pick 3 out of 8, and that is \(\dbinom{8}{3}=\frac{8\times 7\times 6}{3\times 2}=8\times 7=54\)
that will be the denominator
then for the numerator you take the number of ways to pick the three strongest out of the 8
there is only one way to do that
your answer using this method is
\[\frac{1}{54}\]

Answer 12

56 haha

Answer 13

8x7=56
but thanks i think i have it now

Answer 14

the other way is to compute as follows
probability she picks one of the strongest as first pick is \(\frac{3}{8}\) then probilility the next pick is from the strongest 3 is \(\frac{2}{7}\) and finally the last pick has probability \(\frac{1}{6}\) multiply these together and you get
\[\frac{3}{8}\times \frac{2}{7}\times \frac{1}{6}=\frac{1}{\text{whatever eight times seven}}\] is

Answer 15

either method works, and it requires not thinking "do i use permutations or combinations" but rather understanding what exactly the question is asking
also helps to know \(8\times 7\)

Answer 16

Alright man thanks a bunch but....
un mas?

Answer 17

post

Answer 18

There are 7 books numbered 1-7 on the summer reading list. Peter randomly chooses 2 books. What is the probability that peter chooses books numbered 1 and 2?

Answer 19

similar to last one

Answer 20

picks one or two on first choice with probability \(\frac{2}{7}\) then picks the other (given that fist pick was either one or two) with probability \(\frac{1}{6}\)
multiply them together because this is an "and" statement and get
\[\frac{2}{7}\times \frac{1}{6}\] whatever that is

Answer 21

other method
you have \(\dbinom{7}{2}=\frac{7\times 6}{2}\) ways to pick any two from a set of 7, and only one way to pick both one and two in two picks
answer is
\[\frac{1}{21}\] for either method

Answer 22

ah i see,
but how does the and statement change it, what if it was "or" instead of "and"?

Answer 23

depends on the context, but sometimes "or" means "add" or rather add and subtract the intersection
it is hard to answer a general question like that because english language is not math, so it is ambiguous