arranged

unique · 2006-10-27 04:16:06

6 law books and 2 math books sit on the library shelf
how many differnet ways can they be arranged if teh two different types of books are kept together?

i say 12
because 2 times 6 equals 12

Devantè · 2006-10-27 04:45:09

I'm confused as to what you're asking.

So, let the lawbooks be a, b, c, d, e and f

Let the math books be g and h

Do you mean to arrange them like this:

abcdefgh
bacdefgh
cabdefgh
dabcefgh
eabcdfgh
fabcdegh

etc.

Are you supposed to keep all the books in 2 sections (one for law and one for math), and have the 2 sections put together?

If not, 40320, which is 8!. That is, if the math books could be mixed up anywhere in-between the lawbooks.

Last edited by Devanté (2006-10-27 04:47:03)

unique · 2006-10-27 05:06:09

ok like for teh way we did this :

http://www.mathsisfun.com/forum/viewtopic.php?id=4679

cant we do teh same way with this question?

Devantè · 2006-10-27 05:11:20

So, how many different ways can the books be arranged?

Then it is 40320, which is the same as 8!.

unique · 2006-10-27 05:24:31

ok so its almost the same way like we did for the last problem. right

Ricky · 2006-10-27 06:05:45

No, you're interpretting the question wrong. Books are only "different" if they are on different subjects. That is, a math book is considered the same as another math book, even if they are different math books.

So here is the way I would approach this:

Represent law books by 0 and the math books by 1.
Place all the law books on the book shelf.

0 0 0 0 0 0

Now we wish to *insert* the two math books. So take the first one. Where can we insert it?

_0_0_0_0_0_0_

It can go on any one of the '_', and we will get a different sequence.

This problem is exactly the same as the "balls in boxes" problem. We which to put 2 identical "balls" (math books) into 7 identical "boxes" (positions on the bookshelf).

The forumula for this is:

C(n + r - 1, r) = C(7 + 2 - 1, 2) = C(8, 2)

Let me know if you want a combinatorial argument of the forumla. It may take a while to type, so I will only do so if you request.

unique · 2006-10-27 06:20:26

please do explain more

pi man · 2006-10-27 08:05:00

The way I read the orginal question is that books on the same subject must be grouped together. So you could have the 2 math books sitting to the left of the law books or the 2 math books sitting to the right of the law books:

MM LLLLLL or LLLLLL MM.

There are only 2 ways to arrange the math books: M1 M2 or M2 M1. Thats 2!, by the way.
There are 6! ways to arrange the law books. That's 720.
You have to multiply these 2 numbers (2 and 720) together to get the numbers of ways to arrange them in MMLLLLLL fashion. That's 1440 ways.

There are an equal number of ways to arrange them in LLLLLLMM fashion. So multiply the 1440 by 2 to get a total of 2880. A

Ricky · 2006-10-27 08:33:30

Upon further investigation, pi man has the correct interpretation and the correct answer.

unique · 2006-11-01 10:37:47

ok this one

How many distinguishable permutations can be formed in the letters of the word " success"?

i got 7

unique · 2006-11-01 10:44:59

success
uccesss
ccesssu
cesssuc
esssucc
sssucce
ssucces

Ricky · 2006-11-01 13:31:17

There are seven slots we must add:

3 s's
2 c's
1 u
1 e

The order we chose doesn't matter, so let's just start with the s's.

We need to choose 3 slots for the s's:

C(7, 3)

Now we need to choose 2 slots for the c's:

C(4, 2)

Now we choose 1 slot for the u

C(2, 1)

Now we choose 1 slot for the e:

C(1, 1)

All of these events are independant, meaning we can multiply them all together to get the total:

C(7, 3)C(4,2)C(2,1)C(1,1)

pi man · 2006-11-01 17:41:51

I'm not sure if this will result in the same answer as Ricky's but here's some general formulas.

The number of ways to arrange 7 unique letters (abcdefg) is 7!

For each duplicate letter, divide by 2!. abcdeff gives you 7! / 2! permutations and abcddee gives you 7! / (2! * 2!).

For a triplicate (don't know if that's a word!), divide by 3!. So abcdeee gives you 7! / 3! permutations.

SUCCESS is a seven letter word with one duplicate and one triplicate. So you have 7! / (3!*2!) permutations. Thats 420. And that does give you the same results as Ricky.

unique · 2006-11-02 03:00:57

thank you very much.
i understood ricky's way but your way seems direct , i understood a bit more.
thank you both

Math Is Fun Forum

#1 2006-10-27 04:16:06

arranged

#2 2006-10-27 04:45:09

Re: arranged

#3 2006-10-27 05:06:09

Re: arranged

#4 2006-10-27 05:11:20

Re: arranged

#5 2006-10-27 05:24:31

Re: arranged

#6 2006-10-27 06:05:45

Re: arranged

#7 2006-10-27 06:20:26

Re: arranged

#8 2006-10-27 08:05:00

Re: arranged

#9 2006-10-27 08:33:30

Re: arranged

#10 2006-11-01 10:37:47

Re: arranged

#11 2006-11-01 10:44:59

Re: arranged

#12 2006-11-01 13:31:17

Re: arranged

#13 2006-11-01 17:41:51

Re: arranged

#14 2006-11-02 03:00:57

Re: arranged

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