Question

In how many ways can 5 people be seated on a sofa,If there are only 3 seats available.

Answer 1

This is how you do it. Since there are 3 seats available and only 5 people you multiply like this: 5*4*3 = ?. You go down 1 number for each time you multiply, so there are no duplicates, and only multiply like that for 3 times, because that is the number of seats available.

Answer 2

I got half of what you said.

Answer 3

Combinations without repetition (n=5, r=3)
Using the first 5 items: {a,b,c,d,e}

List has 10 entries.
{a,b,c} {a,b,d} {a,b,e} {a,c,d} {a,c,e} {a,d,e} {b,c,d} {b,c,e} {b,d,e} {c,d,e}

Answer 4

How @nincompoop

Answer 5

5P3 = 5!/(5-3)! = 5*4*3 = 60 ways

Answer 6

60 is the answer right?

Answer 7

Yes

Answer 8

Uri this should go to math.

Answer 9

Not history.

Answer 10

\[\frac{ n! }{ \left( n-r \right)!\left( r \right)! }\]

Answer 11

I got it,I did that Ques ^10P3 it is done in same way,I got it..Wait nin you're doing in the method of 5C3?

Answer 12

5C3 =5P3/3!

Answer 13

Omg i posted in History,Dahell i didn't even see it :o @selrachcw95

Answer 14

that is without repetition

Answer 15

So we have to do it in which method..? The one @mangorox and @Nurali said right? :)

Answer 16

Np just close it and move to math. :D

Answer 17

now i'm between and it's almost done. @selrachcw95

Answer 18

I'm in*

Answer 19

if you want the order to matter then 60
using
\[\frac{ n! }{ (n-r)! }\]

Answer 20

ok since I'm extremely nice won't report u. :D

Answer 21

Permutations without repetition (n=5, r=3)
Using the first 5 items: {a,b,c,d,e}

List has 60 entries.
{a,b,c} {a,b,d} {a,b,e} {a,c,b} {a,c,d} {a,c,e} {a,d,b} {a,d,c} {a,d,e} {a,e,b} {a,e,c} {a,e,d} {b,a,c} {b,a,d} {b,a,e} {b,c,a} {b,c,d} {b,c,e} {b,d,a} {b,d,c} {b,d,e} {b,e,a} {b,e,c} {b,e,d} {c,a,b} {c,a,d} {c,a,e} {c,b,a} {c,b,d} {c,b,e} {c,d,a} {c,d,b} {c,d,e} {c,e,a} {c,e,b} {c,e,d} {d,a,b} {d,a,c} {d,a,e} {d,b,a} {d,b,c} {d,b,e} {d,c,a} {d,c,b} {d,c,e} {d,e,a} {d,e,b} {d,e,c} {e,a,b} {e,a,c} {e,a,d} {e,b,a} {e,b,c} {e,b,d} {e,c,a} {e,c,b} {e,c,d} {e,d,a} {e,d,b} {e,d,c}

Answer 22

Okay we can do it in both ways? @nincompoop :D

Answer 23

Just remember where u are next time. :D

Answer 24

@selrachcw95 Thanks :D but i didn't do it puposely :3

Answer 25

Ok. :D

Answer 26

I barley report anyone.Lol!!!

Answer 27

sometimes there are some rules or constraints that may be imposed like can we repeat numbers or should the order of number matter
the first one I gave you cannot have the elements in the set repeated and the order didn't matter. so the formula is different.

Answer 28

@selrachcw95 just stop typing. you aren't even helping
are you drunk or something?

Answer 29

oldrin is typing, now I am scared LUL

Answer 30

if you want numbers, this might be better
Permutations without repetition (n=5, r=3)

List has 60 entries.
{1,2,3} {1,2,4} {1,2,5} {1,3,2} {1,3,4} {1,3,5} {1,4,2} {1,4,3} {1,4,5} {1,5,2} {1,5,3} {1,5,4} {2,1,3} {2,1,4} {2,1,5} {2,3,1} {2,3,4} {2,3,5} {2,4,1} {2,4,3} {2,4,5} {2,5,1} {2,5,3} {2,5,4} {3,1,2} {3,1,4} {3,1,5} {3,2,1} {3,2,4} {3,2,5} {3,4,1} {3,4,2} {3,4,5} {3,5,1} {3,5,2} {3,5,4} {4,1,2} {4,1,3} {4,1,5} {4,2,1} {4,2,3} {4,2,5} {4,3,1} {4,3,2} {4,3,5} {4,5,1} {4,5,2} {4,5,3} {5,1,2} {5,1,3} {5,1,4} {5,2,1} {5,2,3} {5,2,4} {5,3,1} {5,3,2} {5,3,4} {5,4,1} {5,4,2} {5,4,3}

Answer 31

In how many ways can 5 people be seated on a sofa,If there are only 3 seats available.

You want the number of ways that you can *permute* the people among 3 seats... consider each seat separately: