I've got a problem where there are 6 seats in a row, with 6 people to be seated. I need to figure out all possible seating arrangements, given that two people are married and are to be seated next to each other.
Unless I am mistaken, a full seating arrangement list without any conditions would simply be 6! = 720 possible ways of seating everyone (this equates to a permutation of 6 choosing 6, or 6!/(6-6)!).
However, with two specific people out of the 6 needing to be next to each other, that would make for 10 possible places these two could be seated in the row. I can do this by basic counting of spaces (draw out an array of squares and counting the forward and reverse possibilities of adjacent pairs in the set), but my first problem here is I cannot see a logical calculation for this from the given conditions. I'd expect it to be some sort of permutation of two out of the six spaces, but for that I get 6!/(6-2)! = 30, which is the ways of seating any two people out of 6 chosen and not just the couple…so that's not the proper approach. Using a permutation of n=2 also does not work, and does not seem logical.
After this issue, my approach is to multiply this "10" value into the possible placements for the remaining people, which I'd expect to be 4!, given that there are four people and four spaces in which to put them, so as with the full arrangement list, this would be 4!/(4-4)! = 24.
This gives me 10*4! = 240 possible ways of seating everyone; however, this is not the right answer (the correct answer is 48).
Best Answer
We have persons $a,b,c,d,e,f$ with persons $a$ and $b$ being married and needing to sit next to one another (and no other married couples needing to be sat together). We want to count the number of ways these six people can be sat along a row (with a distinct start and end)
For the moment, denote the married couple $a,b$ as a single entity: $X$. How many ways do we have to arrange $X,c,d,e,f$ in a row? (e.g. one such outcome could look like $cXefd$)
Now that we have arranged the "five", replace $X$ with $a$ and $b$ again, but decide how to arrange $a$ and $b$ when putting it back in. (e.g. from $cXefd$ we now make it either $cabefd$ or $cbaefd$)
By multiplication principle, the total number of arrangements is the multiplication of the number of choices at each of these two steps for a total of:
If instead there are multiple married couples, each of which need to sit next to eachother then do similarly. Suppose $a$ and $b$ are married and need to sit next to eachother, $c$ and $d$ are married and need to sit next to eachother, and $e$ and $f$ are married and need to sit next to eachother.
For the moment, treat $a$ and $b$ as one collective entity: $X$. treat $c$ and $d$ as one collective entity: $Y$. treat $e$ and $f$ as one collective entity: $Z$.
How many ways are there to arrange $X,Y,Z$? (e.g. $YXZ$)
Replacing $X$ with $a$ and $b$, how many ways can you do that? (e.g. $YbaZ$)
Replacing $Y$ with $c$ and $d$, how many ways can you do that? (e.g. $cdbaZ$)
Similarly, replace $Z$ with $e$ and $f$, in how many ways? (e.g. $cdbaef$)
By multiplication principle, multiply the number of ways of accomplishing each step for a total of:
Given your statement in your post that this was the answer you expected, it appears that this was the interpretation and that you had neglected to mention that there are multiple married couples, all of which must sit near one another.