In how many ways can 3 singers be selected from 5 who came to an audition?
5 • 4 • 3 = 60
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i dont get how you got the 4 and 3
Okay you have 5 singer yes?
and you have only 3 sets
5 singers, 3 places. 5 options for space#1. 4 left for place #2. and 3 left for place #3
yes thats correct @xapproachesinfinity
cause you can't pick a person twice. make sense?
Yes! like @StudyGurl14 said once the seats are finished we stop counting because we got what we want
oh okay i got it now thx :)
Its just from 5 singers you are choosing 3 of them, and order does not matter. so we do the operation 5C3
No its permutation! 5P3
No it should combinations, because we dont know which one could sing or not, so order does not matter, we can choose any of the 5 people.
@ganeshie8
that doesn't matter we don't have labels we singer 1 for set one singer two for seat 2 and singer 3 for 3 we have 2 singer left that the same as \(\Large \frac{5!}{(5-3)!}\)
i feel it should be a combination because of the keyword "selected"
it could be a permutation also if we are ordering them for singing in the contest
well that's permutation we select the first singers to arrive and count
the total of counting 5 singers is 5! then we need to take out 2! from the total since we have only 3 seats
selecting "mary, abby, ashley" should not be different from selecting "abby, mary, ashley" since the question doesn't talk about first,second,third places right ?
if we need to arrange them seats, it becomes a permutation however
but the question as it stands... im inclining more towards comb
eh i agree with the example you give for this question i believe it's permutation lol
eh okay i changed my mind! after more thinking comb is what fits this
:) i still feel uneasy though because of the work "audition" audition usually refers to ranking.. thats the reason permutation comes to mind first, but again the question doesn't talk about prizes... so...
hehe! you made feel unsure as well
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