This is a problem from 'Introduction to Counting and Probability' from the Art of Problem Solving series.
The Smith family has 4 sons and 3 daughters. In how many ways can they be seated in a row of 7 chairs such that at least 2 boys are next to each other?
Here's book's answer for clarification:
According to the book, the only way to assign genders to the seating so that no two boys are next to each other is BGBGBGB. However, BGBGBGB is not considered to be a single seating arrangement but rather the set of 144 (i.e. 4! * 3!) seating arrangements, where there are 4 boys, 3 girls, and 7 seats.
From the observer's perspective, why aren't all 4 boys considered to be indistinguishable from each other (the same goes for all 3 girls), since we are only concerned about the gender? Why do we consider BGBGBGB to be the set of 144 seating arrangements? It seems to me that there is only one distinct ordering such that two boys are not sitting next to each other.
$\endgroup$ 22 Answers
$\begingroup$Math can tell us how to answer questions, but not how to ask them. "Do we want to distinguish the individual children?" is a matter of what you want to know. (Though in a textbook problem, I guess it's an issue of what the author of the textbook intended you to want.) If the Smith family is asking you to describe a seating arrangement, do you want to tell them "put two boys first, then three girls, then the last two boys" or "put Tom, then Harry, then Anna, then..."?
The thing that is important mathematically, and not just personal choice, is consistency. In this problem, we're employing complementary counting to subtract from all $7!$ seating arrangements the ones where no two boys are seated together. The value $7!$ distinguishes between all the children; therefore the value we subtract must also distinguish between the children. Computing $7! - 144$ is valid; computing $7! - 1$ is not.
We could also be consistent and distinguish between the children in neither case. In that case, there are $\binom{7}{3} = 35$ total seating arrangements (BBBBGGG, BBBGBGG, BBBGGBG, ..., GGGBBBB). Of these, exactly $1$ (BGBGBGB) does not have two boys sitting next to each other, so there are $34$ that do.
We could even first count the boy/girl classes of seating arrangements and get $34$, and then add in the fact that for each of them, we have $4! \cdot 3! = 144$ ways to decide which boys are which and which girls are which. This gives $34 \cdot 144 = 4896$ as the answer, the same as the textbook method.
$\endgroup$ 4 $\begingroup$The BGBGBGB arrangement is a restriction of the possibilities, but the answer isn't just "BGBGBGB", because we care about details within that restriction.
Suppose we take your line of thinking but remove the restriction to BGBGBGB. In other words, consider the problem, "How many ways can $4$ boys and $3$ girls be seated in a row?" Your reasoning would give $1$ (you just line them up), which is not what is intended by the question.
To look at it from another point of view, consider that human beings are unique. It might matter a lot which people sit where. This is different from having $4$ black pearls and $3$ white pearls and placing them on a string so that no two adjacent pearls are the same colour. In that case, the answer is indeed $1$.
$\endgroup$ 3
