# permuattions question - paraphrased...

Discussion in 'Mathematics' started by beedge, May 16, 2011.

1. ### beedgeNew commenter

You have n boys sitting on a bench.
2 of them can not sit at the end.
A different 2 must sit together
How many ways are there of achieving this?

I was quite pleased with coming up with 2(n-2)(n-3)(n-3)! in a few minutes, until I found out it was wrong.
This was more-or-less my [flawed] reasoning, which I hope makes sense:
Let's consider the 2 boys who can't sit in the middle. There are n-2 positions the first one can go in. The second can go in n-3 positions. So far that's (n-2)(n-3) ways of arranging those 2.
We now have n-2 boys left to place. But two of them have to sit together - so let's call those one unit - so we consider how to place n-3 objects. So that's (n-3)! ways (x 2 of course to allow for the congealed boys to swap places).
So putting it altogether, you get 2(n-2)(n-3)(n-3)!
Which is wrong.
The problem is, after I'd thought about it, there aren't (n-3)! x 2 ways of arranging the final n-2 boys. There would be if there were n-2 empty places next to each other, but that of course is not the case.
And this is the whole problem - depending on WHERE the first 2 boys go, affects how many ways you can position the remaining n-2 boys. So you can't say for every x positions the first 2 go in, there are y others for the remaining n-2, therefore the answer is XY.
I spent hours on this, and still can't get it.
Yet I can't have been that far out in the first place,as the correct answer is:
2(n-3)(n-4)(n-3)!
which at least looks like my first attempt.
Please can you explain how to get this.
Thanks very much.

2. ### PiranhaStar commenter

Think about it as n-1 boys, and double it at the end, as you point out. I think it is easier to think about who <u>can</u> sit at the ends.
Consider the end two places. There are n-3 candidates for the left one, and n-4 for the right. So, (n-3)(n-4).
Now, the other n-3 can be arranged in (n-3)! ways. Double it for swapping the pair round and you get (n-3)(n-4)(n-3)!

3. ### PiranhaStar commenter

Or even 2(n-3)(n-4)(n-3)!

4. ### NazardNew commenter

Treat the "2 who must sit together" as a single entity - you therefore have n-1 entities.
(n-3) of them can sit in the first position (because two of the n-1 can't)
(n-4) of them can then sit in the last position.
This leaves the middle ones, which is (n-3)!
Finally, the chums can sit either way round so this gives a factor of 2.
Resulting in: 2(n-3)(n-4)(n-3)!

6. ### David GetlingSenior commenter

beedge, that was quite a nice question. Where did it come from? There was a similar one on last May's IB Higher Level paper 2.

7. ### beedgeNew commenter

It actually came an algebra book called "Advanced Algebra" that my Dad gave me. He was given it at school in the 60's when doing A-level.
Of course the gulf in standard between the questions in this book, and the questions in modern A-level is just ridiculous. But I guess we knew that already.
Anyway, did you mean this question?<font face="TimesNewRomanPSMT"> "Three Mathematics books, five English books, four Science books and a dictionary are
</font>to be placed on a student&rsquo;s shelf so that the books of each subject remain together.
(a) In how many different ways can the books be arranged?
<font face="TimesNewRomanPSMT">(b) In how many of these will the dictionary be next to the Mathematics books?"

I'm still trying to master this topic, so please correct me, but is this the answer?
(a) consider the 3 Maths books, the 5 English books and the 4 Science books as individual objects. Plus the dictionary makes 4.
So we have 4! ways of arranging them.
The 3 maths books can be arranged in 3!
The 5 English books in 5!
The 4 English books in 4!
So we have 4! x 3! x 5! x 4! in total
Assuming it's not a circular bookshelf !
Initially, I've ignored the dictionary.
So therefore we have 3! x 3! x 5! x 4! ways of arranging the other books. Now for every one of those positions, the dictionary can go before OR after the maths books, i.e 2 positions.
Therefore 2 x 3! x 3! x 5! x 4!

</font>

8. ### David GetlingSenior commenter

Yes, beedge, that's the question, and your reasoning is correct - or else I've just taught this wrongly to a bunch of kids.

Couldn't agree more about the decline in A-level maths. To the best of my memory, everything in C1 to C4 and FP1 to FP3 used to appear in the Pure Maths syllabus, PLUS some stuff on partial differentiation.

As I've said several times before I think IB is better, and if you look at the HL options (all of which, plus an extra geometry one, have to be taken for IB Further Maths) it really beats A-Level maths (at least Edexcel) hands down.