0
$\begingroup$

6 people sit down a round table. 4 of them belong to group X and 2 of them belong to group Y.

How many ways are there for the 6 people to sit down by taking into account that the 2 people in group Y must sit down side-by-side?

And how many ways are there if the 2 people in group Y must sit away from each other?

My Approach:

For the first question I assume that the 2 people in group Y can take places 1 to 6 and permuting both people in this positions there are a total of 6*2 ways. Then the other 4 people can be sited in the other 4 free places, this is 4! = 24. So in total, 12*24 = 288 ways.

For the second question I believe the circular permutation can be used first in order to sit the 2 people in group Y away from each other, this is $$ \binom {6}{2}*(2-1)!= 15$$ and then the other 4 people could be sited, this is 4! = 24. So in total 15*24=360 ways.

Is my approach correct?

$\endgroup$
  • 1
    $\begingroup$ To seat two people away from each other, you have a free choice of the 6 seats for the first person, and 3 seats available thereafter for the second. $\endgroup$ – Joffan Feb 1 '15 at 11:22
1
$\begingroup$

For sitting group $Y$ together:

Seat one of the people in group $Y$ arbitrarily. We count this as being do-able in $1$ way because of the symmetry of the circle.

Seat the other group $Y$ person ($2$ ways).

Seat the remaining $4$ people (group $X$) ($4!=24$ ways).

Total: $48$ seatings possible.

For sitting group $Y$ apart:

Seat one of the group $Y$ people ($1$ way (as above)).

Seat the other group $Y$ person ($3$ ways--to keep that person away from the other $Y$ person).

Seat the $4$ from group $X$ ($24$ ways).

Total: $72$ seatings possible.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.