circular perms Q (1 Viewer)

[jst.KP]

There are two distinct round tables, each with five seats. In how many ways may a group of 10 peoplebe seated?

 

[lychnobity]

There are two distinct round tables, each with five seats. In how many ways may a group of 10 peoplebe seated?

Answer: 2C1 x 10C5 x 4! x 2

Explanation: 2C1 to choose a table, 10C5 to choose the 5 people on that table, 4! to arrange them, and 2 because they can swap tables (ie say they chose table A at first, and now all those people want to go to table B)

 

[Timothy.Siu]

9P4x4!
maybe

 

[Drongoski]

There are two distinct round tables, each with five seats. In how many ways may a group of 10 peoplebe seated?

Is answer: 2 x {10C5 x 4! + 4! } ??

prob making a fool of myself!

Edit: ought to be 2 x {10C5 x 4! x 4!}

Perhaps still wrong

 

[lychnobity]

Is answer: 2 x {10C5 x 4! + 4! } ??

prob making a fool of myself!

In regard to this part: 10C5 x 4! + 4!

Why did you add 4! ?

 

[Drongoski]

I thought after filling table-1, u have 5 persons left who can fill table-2 in (5-1)! ways. u double swapping tables. But I'm always nervous doing combinatorics; easy to overlook something.

Edit:

lychnobity

when u alerted me it stilll didn't dawn on me '+' should have been 'x'

 

[CountVonChocula]

u minus 1 if it is a short and circular conversation

 

[jst.KP]

the answer is 145 152 if that helps btw

 

[oly1991]

9P4x4!
maybe

the answer is 145 152 if that helps btw

tim has the right answer except he has to times it by 2

 

[Timothy.Siu]

tim has the right answer except he has to times it by 2

hmm it could be that.

or i did a mistake in my calculation,
can someone answer this, if u have 10 guys and ur seating 5 of those on a table wats the number?
is it 10x9x8x7x6/5 ? because if it is, thats wat i did wrong

 

[gurmies]

hmm it could be that.

or i did a mistake in my calculation,
can someone answer this, if u have 10 guys and ur seating 5 of those on a table wats the number?
is it 10x9x8x7x6/5 ? because if it is, thats wat i did wrong

10C5 x 4!

Selecting 5 guys from a possible 10 and then arranging around a table (5-1)!...Same as what you got there Tim..

 

[undalay]

10C5x4P4x4P4

From 10 make 2 distinct groups, the re-arrange each table.

On the right track?

 

[Timothy.Siu]

10C5x4P4x4P4

From 10 make 2 distinct groups, the re-arrange each table.

On the right track?

looking good

 

[micuzzo]

tim has the right answer except he has to times it by 2

thats right... 9P4 x 4! x 2

 

[gurmies]

10C5 x 4! x 4! was my expression :S Still the same answer, but is my reasoning legitimate? 10C5 ways to choose 5 people from 10, then 4! ways to arrange each at their table...

 

[micuzzo]

10C5 x 4! x 4! was my expression :S Still the same answer, but is my reasoning legitimate? 10C5 ways to choose 5 people from 10, then 4! ways to arrange each at their table...

well im not sure... the question is dumb... firstly i would say that since its a circle u need to minus 1 ie (n-1)! so i wouldnt use 10

then it says they are distinct ... so i guess u can assume that it will be a permutation... but wats a distinct table, duz it mean that the tables are diff from each other or that they are the same but seating arrangements are diff

therefore

9P4 (for the table) x 4! (bcoz they can be arranged in 4 diff ways on the table) and x 2 (because there are two tables)

I think thats the right way... not sure though...

 

[oly1991]

well im not sure... the question is dumb... firstly i would say that since its a circle u need to minus 1 ie (n-1)! so i wouldnt use 10

then it says they are distinct ... so i guess u can assume that it will be a permutation... but wats a distinct table, duz it mean that the tables are diff from each other or that they are the same but seating arrangements are diff

therefore

9P4 (for the table) x 4! (bcoz they can be arranged in 4 diff ways on the table) and x 2 (because there are two tables)

I think thats the right way... not sure though...

yeh thats what i thought it was

 

[khorne]

yeh thats what i thought it was

This is fine, but, as probably mentioned before:

10C5 ( choose 5 people) x 4! (arrange them) x 4! (arrange remaining 5)

so 252 x 4! x 4! = 145 152

However, guys, I don't agree with the method above, the permutations one...

I would do it as, 10! (each person can sit in each seat) however, because of the circles, there is one true position for every 5 cyclic positions, and there are tow tables, so 10!/5 and then /5 again (or 10!/25)

Personally the combinations one makes more sense to me, but whatever floats your boat.