Table of Friends
Probability
Level
3
A group of 8 friends sit together in a circular table. If Marie refuses to sit beside Manny unless Mary sits on the other side of Manny as well, how many possible seating arrangements are possible?!
The answer is 3840.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Total ways to arrange 8 people around the table is (8-1)!=7!;
Let's denote Marie as "A", Manny as "B" and Mary as "C" to prevent confusion.
Since we do not want A to sit next to B, let’s try and make them sit together. This will give us the number of arrangements that are unacceptable to us. Let’s say that A and B are a single unit. So now there are 7 units which need to be arranged in a circle. This can be done in (7-1)! = 6! ways. Since there are two arrangements possible, AB and BA, within the unit, we need to multiply 6! by 2.
Number of arrangements in which A and B sit together = 2 * 6!
We can subtract these ‘unacceptable arrangements’ from total arrangements to get the number of ‘acceptable arrangements’. But this number of ‘unacceptable arrangements’ includes those arrangements where C is sitting on the other side of A. But those arrangements are acceptable to us so we should not subtract them out. How many such arrangements are there in which A and B are sitting together and C is sitting beside A too?
Now C, A and B form a single unit leaving us with 6 units to be arranged in a circle. 6 units can be arranged in (6-1)! = 5! ways
CAB can also be arranged as BAC, hence the 5! needs to be multiplied by 2. (Mind you, we will not consider ABC, ACB etc here since A should be in the middle)
Number of arrangements in which A and B sit together and C sits beside A = 2 * 5!
Therefore, number of unacceptable arrangements = 2 * 6! – 2 * 5!
We subtract these out of the total number of arrangements and we get the total number of acceptable arrangements.
Possible number of seating arrangements = 7! – (2 * 6! – 2 * 5!) = 3840
If you are wondering about the ‘painful’ calculation involved in the step above, don’t worry. Calculations with factorials are generally quite straight forward.
7! – (2 * 6! – 2 * 5!) = 7! – 2 * 6! + 2 * 5!
= 2 * 5! (21 – 6 + 1) (Take 2 * 5! common out of the three terms)
= 2 * 120 * 16 = 32 * 120 = 3840