A Strange Question

Can someone help explain the theory behind the bottom question or give me a hint without giving me the answer. It's a bonus question, but I rather get assistance before copying someone completely.

I already know Permutations and Combinations and the Fundamental Counting Principal, I'm just uncertain of how to apply them to the question. Also, with the numbers being so large I can't write out all the combinations to try and figure out the question that way either.

In how many ways can 6 girls sit in 10 chairs if sue and sally must sit together?

Thanks,

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Pyrrhus
Homework Helper
Well first you can calculate the possibility the other girls will sit, then you know 2 of the must be together, so you can calculate the different places those two girls will pick.

$$_{n}P_{r} = \frac{n!}{(n-r)!}$$

$$_{10}P_{4} = \frac{10!}{(6)!}$$

$$_{2}P_{2} = \frac{2!}{(0)!}$$

$$_{2}P_{2}*_{10}P_{4} = 10080 ways$$

I think u missed a factor ....

P(10,4) << number of ways 4 girls can get arranged in 10 seats

6 seats remain ....
since sally and sue sit together we can club two seats together and say there are 5 seats of which we need to choose one and we can do that in
P(5,1)

and ofcourse sally and sue can be arranged amongst themselves in
P(2,2)
-- AI

Thanks, I'm still a bit lost. 2 P 2 = 2
This is because you can either have 2 people sit or 1 person sit?

Would 10 P 4 x 6 P 2 work? Final answer = 151200?

Hmm, I'll have to look over both your answers some more until I figure it out.

Sue and Sally are supposed to be intpreted as sitting side by side rather than in the same seat, correct?

Last edited:
Pyrrhus
Homework Helper
It would not work, because 2 girls can pick 6 chairs, but that doesn't mean they will sit together, by the way Tenali is right i missed a factor.

Pyrrhus
Homework Helper
Side by side, yes

Thanks - this question has been stumping me. I'm still unsure of which way to go with it.

If sue and sally are to be beside each other they I figured that there are 18 combinations that allow this.

If you use SS12345678 to represent the seats. S-S could also be rearranged to work for Sally, Sue and Sue, Sally?

tony873004
Gold Member
Isn't something missing here?

If you label the seats 1-10, and they are in a circle, then seats 1 & 10 qualify as next to each other. But if they are in a line, they do not.

Pyrrhus
Homework Helper
where does it state they are sitting in a circle?

They are in a line, we haven't been working with circles.

tony873004
Gold Member
How many seat pairs are there that are next to eachother? Let's list them:

1) 1-2
2) 2-3
3) 3-4
4) 4-5
5) 5-6
6) 6-7
7) 7-8
8) 8-9
9) 9-10

So there are 9 seat pairs. If Sally and Sue must sit together, but it doesn't matter in what order, they can sit as Sally & Sue, or Sue and Sally. So for each pair of seats, there's two ways they can sit.

So multiply 9 * 2 for 18 different ways Sally and Sue can sit and still be next to each other.

Now, for each of the 18 different ways Sally and Sue can sit next to each other, how many possibilities are there for the other girls?

There are 8 remaining seats for the 4 remaining girls. Girl #1 has 8 possibilities. Once she is seated, Girl #2 has 7 possibilities. Once she is seated, Girl #3 has 6 possibilities. Once she is seated, Girl #4 has 5 possibilities.

8*7*6*5 = 1680 possible ways for the 4 remaining girls to fill the 8 remaining seats for each of the 18 acceptable pairings of Sue & Sally.

1680 * 18 = 30240 combinations.

Thanks everyone. The last answer given was the one my teacher wanted, if anyone was wondering. Thanks again.

Cyclovenom and I made the same obvious mistake :p
(i wont iterate what our error was .... which is unnecessary anyways)

Tony is right with his solution but i still have a small nitpicking to do :D

So there are 9 seat pairs. If Sally and Sue must sit together, but it doesn't matter in what order, they can sit as Sally & Sue, or Sue and Sally. So for each pair of seats, there's two ways they can sit.

So multiply 9 * 2 for 18 different ways Sally and Sue can sit and still be next to each other.
Well here the order in which they sit is important hence u multiply by 2 ... if the order were not important then you wouldn't have multiplied by 2 ...

Sorry i couldn't help nitpicking :)
But i didn't want a small hole to be there in the solution which is otherwise perfect ...

-- AI