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Permutation and combinations

Punch

New member
Jan 29, 2012
23
Four married couples attend a wedding dinner. One of the couples brought along two children. Find the number of ways in which these ten people can be seated round a table if each couple must sit together.

I need to know the logic and thinking process behind how the answer is derived.

What I tried is:
First person has 10 seats to choose, second person 8 seats to choose and so on. Each couple can then seat on different sides.

10(8)(6)(4)(2^5)=61440

Correct answer is 1920

Another way of thinking I have is this: Consider each couple and the 2 children as each individual groups.

Total number of ways of arranging the 5 groups in a round table is (5-1)!=24

Then permutate each couple and children=2^5

so total number of ways = (2^5)24=768
 
Last edited:

ThePerfectHacker

Well-known member
Jan 26, 2012
236
Hello,

Imagine a round table with ten positions open. Where can the first kid be seated? Anywhere, he can sit anywhere he pleases to. How many choices does he have? 10.

Now where can the second kid be seated? Draw drawining some pictures here, but you should realize that the second kid cannot be sitting 1 seat apart from the first kid. Because there is no room for couples to sit together there! Also, the second kid cannot be sitting 3 seats apart from the first kid for the same reason. You should see that the second kid can only be sitting an even number of seats away from the first kid. Thus, the second kid has only 5 choices.

Start with the first kid. Move over to the next avaliable seat (in a clockwise manner). How many people can sit there? There are 8 remaining people and so there are 8 choices avaliable. But in the next avaliable seat who can sit there? It must be the spouse, which means there is only 1 choice, he/she is forced into that seat.

Move to the next avaliable seat. How many people can sit there? Now there are 6 people remaining and so there are 6 choices avaliable. But in the next avaliable seat it must be the spouse, so he/she is forced into that seat.

Move to the next avaliable set. Same reasoning tells us that there are 4 people remaining, and so 4 choices.

Finally with the last two seats remaining next to eachother there is just 2 ways to seat those couples together.

Thus, we get 10*5*8*6*4*2 = 19200.

So how do you get an answer of 1920? I guess it is because in your problem no person is designated as the "head" of the table, i.e. a rotation of all people one seat over is considered to be the same seating arrangement. As there are 10 rotations in seating the answer without any head of table needs to be divided by 10. That is how they got 1920.
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
Four married couples attend a wedding dinner. One of the couples brought along two children. Find the number of ways in which these ten people can be seated round a table if each couple must sit together. I need to know the logic and thinking process behind how the answer is derived.
There are always multiple ways to do these. Here is another.
There are six units: four couples and two children.
Seat any couple at the table together. It is now an ordered table.
There are $5!$ ways to seat the remaining five units.
But each couple can be seated in two ways.
Thus $5!\cdot 2^4=1920$.