Answer: 1,693,440
This is one single number. The commas are there to make it more readable. There's a possibility your teacher wants you to erase the commas.
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Explanation:
Let's label the seats as [tex]S_1, S_2, S_3, \ldots S_{10}[/tex] from left to right.
Seats [tex]S_1[/tex] and [tex]S_{10}[/tex] are at the left-most and right-most endpoints.
For either seat, we can only choose a boy here. Let's say seat [tex]S_1[/tex] has the first selection of a boy. There are 8 to choose from here. Then seat [tex]S_{10}[/tex] will have 8-1 = 7 choices since no one can be in more than one spot at once.
The remaining seats [tex]S_2, S_3, \ldots, S_9[/tex] will get either a boy or a girl. For now, we'll ignore the requirement that the two girls can't sit together.
Seat [tex]S_2[/tex] has 7-1 = 6 boys and 2 girls to pick from. So we have 6+2 = 8 people to pick from for this seat. Or we could note that if the two endpoints are taken, then 10-2 = 8 choices are left for seat
Seat [tex]S_3[/tex] then has 8-1 = 7 choices
Seat [tex]S_4[/tex] has 7-1 = 6 choices
Seat [tex]S_5[/tex] has 6-1 = 5 choices
and so on until we count down to 1 for seat [tex]S_9[/tex]
To summarize,
- [tex]S_1[/tex] has 8 choices (8 boys)
- [tex]S_{10}[/tex] has 7 choices (8-1 = 7)
- [tex]S_2[/tex] has 8 choices (6 boys+2 girls)
- [tex]S_3[/tex] has 7 choices
- [tex]S_4[/tex] has 6 choices
and so on until we reach [tex]S_9[/tex] having 1 choice.
If we didn't worry about the girls sitting together or not, then we'd have 8*7*8*7*6*5*4*3*2*1 = 2,257,920 different permutations of ten students where neither girl is allowed to sit at the ends of the row.
We'll use this value later, so let A = 2,257,920.
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However, the instructions state that the girls aren't allowed to sit next to each other.
In a strange 180 degree turn, let's consider the case where the girls are only allowed to sit next to each other (i.e. they can't be separated). What I'll do is have them taken out of the lineup and replaced by the teacher.
We go from 10 students to 10-2 = 8 students after the girls are temporarily removed. Then increase that to 8+1 = 9 people to account for the teacher taking the girls' spot. Anywhere the teacher is in this new lineup will represent the girls' location.
Then we'll follow a similar idea as in the previous section.
We have,
[tex]\text{8 choices for seat } S_1[/tex]
[tex]\text{7 choices for seat } S_{9}[/tex] (which is now the right-most seat)
and,
[tex]\text{7 choices for seat } S_2[/tex]
[tex]\text{6 choices for seat } S_3[/tex]
[tex]\text{5 choices for seat } S_4[/tex]
and so on until we get 1 choice for seat [tex]S_8[/tex]
Multiplying those values out gives:
8*7*7*6*5*4*3*2*1 = 282,240
This represents the number of ways to arrange the 8 boys and 1 teacher such that the teacher is not allowed to sit on the ends (otherwise the girls will sit on the ends).
This is almost the number of ways to arrange the 8 boys and 2 girls so that the girls aren't at the ends.
I say "almost" because it's only half the story (quite literally). Wherever the teacher sits, there are two ways to arrange the girls. So if the teacher is say in slot 2, then this could mean we have girlA,girlB for slots 2 and 3, or we could have girlB,girlA for slots 2 and 3. The order is important.
So we'll need to multiply that 282,240 figure by 2 to get the proper count
2*282,240 = 564,480
Now the value 564,480 fully represents the number of ways to arrange the 8 boys and 2 girls in the configuration mentioned above.
Let B = 564,480 because we'll use it later.
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So we did all that work in the previous two sections, and got the results of
A = 2,257,920
B = 564,480
We'll subtract those results to account for cases in which the girls are not allowed to sit together.
A - B = 2,257,920 - 564,480 = 1,693,440
That wraps everything up. This number is bit under 1.7 million.
For a bit of context and comparison, there are 10! = 10*9*8*7*6*5*4*3*2*1 = 3,628,800 ways to arrange the ten students without any extra conditions.
