uri
  • uri
In how many ways can 5 people be seated on a sofa,If there are only 3 seats available.
History
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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.
uri
  • uri
I got half of what you said.
nincompoop
  • nincompoop
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}

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

uri
  • uri
How @nincompoop
Nurali
  • Nurali
5P3 = 5!/(5-3)! = 5*4*3 = 60 ways
uri
  • uri
60 is the answer right?
anonymous
  • anonymous
Yes
selrachcw95
  • selrachcw95
Uri this should go to math.
selrachcw95
  • selrachcw95
Not history.
nincompoop
  • nincompoop
\[\frac{ n! }{ \left( n-r \right)!\left( r \right)! }\]
uri
  • uri
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?
uri
  • uri
5C3 =5P3/3!
uri
  • uri
Omg i posted in History,Dahell i didn't even see it :o @selrachcw95
nincompoop
  • nincompoop
that is without repetition
uri
  • uri
So we have to do it in which method..? The one @mangorox and @Nurali said right? :)
selrachcw95
  • selrachcw95
Np just close it and move to math. :D
uri
  • uri
now i'm between and it's almost done. @selrachcw95
uri
  • uri
I'm in*
nincompoop
  • nincompoop
if you want the order to matter then 60 using \[\frac{ n! }{ (n-r)! }\]
selrachcw95
  • selrachcw95
ok since I'm extremely nice won't report u. :D
nincompoop
  • nincompoop
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}
uri
  • uri
Okay we can do it in both ways? @nincompoop :D
selrachcw95
  • selrachcw95
Just remember where u are next time. :D
uri
  • uri
@selrachcw95 Thanks :D but i didn't do it puposely :3
selrachcw95
  • selrachcw95
Ok. :D
selrachcw95
  • selrachcw95
I barley report anyone.Lol!!!
nincompoop
  • nincompoop
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.
nincompoop
  • nincompoop
@selrachcw95 just stop typing. you aren't even helping are you drunk or something?
nincompoop
  • nincompoop
oldrin is typing, now I am scared LUL
nincompoop
  • nincompoop
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}
anonymous
  • anonymous
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:|dw:1371239992230:dw|
anonymous
  • anonymous
Recall we *multiply* to determine our total number of seating arrangements:$$5\times4\times3=60$$
anonymous
  • anonymous
More generally, the number of partial \(k\)-permutations of \(n\) objects is given by \(_nP_k=\dfrac{n!}{(n-k)!}\)
nincompoop
  • nincompoop
we got that written down already, master jedi
anonymous
  • anonymous
I know, I just wanted to give a little more intuition. It's essentially the same principle at work when computing how many outfits you can make with 3 shirts and 2 p|dw:1371240695188:dw|ants:
selrachcw95
  • selrachcw95
Nincompoop I am not drunk
uri
  • uri
So the answer is 60,First tell me that!! :D

Looking for something else?

Not the answer you are looking for? Search for more explanations.