*Zara*
Lv 4

Mr.Blue, Mr.Black, Mr.Green, Mrs.White, Mrs.Yellow, and Mrs.Red sit around a circular table for a meeting.?

Mr.Black and Mrs.White must not sit together. Calculate the number of different ways these 6 people can sit at the table without Mr.Black, and Mrs. White sitting together.

I worked out this problem and got 576 possible ways as the answer. Can anyone confirm that? If my answer’s wrong, please work it out and show all your work. thank-you so much!

Relevance

Well, let's be clear: a "way of sitting together" is determined by who sits next to whom. So you can represent people sitting together by a string like this:

Be - B - G - W - Y - R

(Be = blue, B = black)

There are 6! strings like this one. However, that string is equivalent to any string like it if you shifted it, since it's a circular table. There are 6 such strings. The same goes for flipping it around (reversing the order). There are 2 such strings. That means that there are actually:

6! / (6 × 2) = 60

-------- -------- ------- -------- -------- ------- -------- -------- -------

To give you an idea of which strings are different and which are the same, consider the following examples, simplified to have only four people seated (R, G, B, Y)

R - G - B - Y

G - B - Y - R

Y - B - G - R

These are all equivalent, because in each of the three arrangements, everyone has the same two neighbors. The second is a "shift" of the first, and the third is a "reverse" of the first. The following two are not equivalent (check the neighbors of R, for example):

G - R - B - Y

B - G - R - Y

This strategy allows the group of 6! rearrangements to be split into equivalent groups of 12, leaving you with exactly 60 non-equivalent arrangements.

-------- -------- ------- -------- -------- ------- -------- -------- -------

Now back to your problem, you need to avoid the strings where white and black sit together. Those can be counted in a completely similar fashion, but you "join" the two together:

Be - [B/W] - G - Y - R

Now you only have to rearrange these five amongst themselves, and B/W amongst themselves, and so in the same fashion as above, this gives us (5!×2!)/(5×2) = 24

So there are 60 - 24 ones that avoid the B/W combination. That's only 36. You have greatly over-counted them, I believe because you did not account for rotation around the table as an equivalent seating position.

The general formula for this is:

n! / (2n) - (n-1)! / (n-1) = n! / 2 - (n-2)!

-------- -------- ------- -------- -------- ------- -------- -------- -------

If you are not allowed to "mirror" or "reverse" the order to obtain an equivalent seating chart, then there are just double 36, which is 72 arrangements. But that makes very little sense, since the two seating arrangements are functionally equivalent.

Okay, this is what I think.

Suppose Blue is already at the table. Then, the rest can be arranged in 5!, or 120 ways.

Now imagine Black and White are already at the table. Treat them as 1 person. Then, the rest can be arranged in 4! ways, or 24 ways. But White and Black can be permutated in 2! ways. Therefore, the 6 can be arranged in 24 * 2! = 48 ways, if Black and White are sitting together.

Therefore, the number of ways the 6 people can sit when Black and White are NOT sitting together = 120 - 48 = 72.

This is if clockwise and anticlockwise arrangements are not considered to be the same.

If they are considered to be the same, then:

Suppose Blue is already at the table. Then the rest can be arranged in 5!/2 ways, or 60 ways.

Now imagine Black and White are already at the table. Treat them as 1 person. Then, the rest can be arranged in 4!/2 ways, or 12 ways. But White and Black can be permutated in 2! ways. Therefore, the 6 can be arranged in 12 * 2! = 24 ways, if Black and White are sitting together.

Therefore, the number of ways the 6 people can sit when Black and White are NOT sitting together = 60 - 24 = 36.

I'm not very very sure, but this is what makes sense to me.

I hope that helps. Forgive me if it's wrong! :)

me07.

• holdm
Lv 7