# Find out the number of ways the people can be seated in a row

I have a question,

There are $4$ boys and $4$ girls and there are $8$ seats. Find the number of ways that they can sit alternatively and certain two of them (a boy and a girl) should never sit together.

I am totally stuck with this one. One way I think I can find the answer is to consider those certain two a single unit and then calculate the seating arrangement ($3!*3!*2$) and then subtract it from the total combinations. Is this the right way to solve such a question?

Thanks

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Let $B_0$ and $G_0$ be the boy and girl who must not sit next to each other. Number the seats $1$ through $8$ from left to right. Suppose that $B_0$ sits in seat $1$. Then there are $3$ girls $-$ all of them except $G_0$ $-$ who can sit in seat $2$. After that the other $3$ boys can fill seats $3,5$, and $7$ in any order, and the other $3$ girls can fill seats $4,6$, and $8$ in any order, for a total of $3\cdot 3!\cdot 3! = 108$ ways to fill the seats with $B_0$ in seat $1$. An exactly similar calculation shows that there are also $108$ ways to fill the seats with $B_0$ in seat $8$.

If $B_0$ chooses to sit in one of the six seats in the middle, matters are a little different, because there are now two seats in which $G_0$ cannot sit. Note, though, that as soon as $B_0$ takes his seat, we know which $4$ seats must be taken by the girls, and we know which $3$ seats must be taken by the other $3$ boys. Those $3$ boys can sit as they please among those $3$ seats, so that’s $3!=6$ choices. $G_0$ has a choice of the $2$ ‘girl’ seats not adjacent to $B_0$: that’s another $2$ choices. Finally, the remaining $3$ girls can sit as they please among the remaining $3$ ‘girl’ seats, for another $3!=6$ choices. The grand total is then $6\cdot 2\cdot 6=72$ ways in which they can be seated once $B_0$ chooses a particular middle seat. Since there are $6$ middle seats, that comes to a total of $6\cdot 72 = 432$ ways.

Finally, combine the two cases: we have altogether $2\cdot 108 + 6\cdot 72 = 216+432=648$ ways to seat the $8$ of them.

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Assumptions:

1. Let’s say that the seats are numbered $1,2,\dots,8$.
2. Let’s say the boys will sit in the odd-numbered seats: $1,3,5,7$. Girls will sit in the even-numbered seats: $2,4,6,8$.

For seat #$1$: you can choose $1$ boy out of $4$ boys in $4$ ways.

For seat #$3$: you can choose $1$ boy out of $3$ remaining boys in $3$ ways.

For seat #$5$: you can choose $1$ boy out of $2$ remaining boys in $2$ ways.

For seat #$7$: you can choose $1$ boy out of $1$ remaining boy in $1$ way.

The number of ways you can seat the boys is $$4! = 4\cdot 3\cdot 2\cdot 1\tag{a}$$

You can do the same for girls to obtain:

The number of ways you can seat the girls is $$4! = 4\cdot 3\cdot 2\cdot 1\tag{b}$$

So, the number of distinct ways to seat the boys AND the girls is the product of $(a)$ and $(b)$:

$$4! \cdot 4!\tag{c}$$

Now, remember our assumptions 1 and 2 above? These assumptions cover only $1$ of $2$ ways we can run the experiment. That is, there is another way in which girls sit in the odd-numbered chairs and boys sit in the even-numbered chairs. The number of ways resulting from this new assumption is exactly the same as the one found in $(c)$ above.

So the total number of ways $=4! \cdot 4! + 4! \cdot 4! = 2 \cdot 4! \cdot 4!\;.$

Edit: As per Brian M. Scott comment below, this solution is incomplete.

We need to subtract the number of ways the given pair can sit together from the number shown above.

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You’ve overlooked the fact that there are a boy $b_0$ and a girl $g_0$ who must not sit next to each other. –  Brian M. Scott Oct 24 '11 at 9:41
@BrianM.Scott, thanks for your comment. You are correct. –  Emmad Kareem Oct 24 '11 at 16:46
$N=8!-2*7!=8*7!-2*7!=6*7!$