Circular permutation, sitting 6 people in a round table

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?

• 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. – Joffan Feb 1 '15 at 11:22

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.