Question

Suppose you want to hang 5 different pairs of socks on a clothesline to dry. How many ways can you arrange the socks on the line?

The answer is 113,400 ways but I'm not too sure how to get this answer! I'd love to know the process and any help is very appreciated :)

Answer 1

Thanks so much for the reply!! I forgot to add that the answer is 113, 400 :)

Answer 2

Maybe, if you check out topics on combinations and permutations, it would help. Take a look at this: http://url.moosaico.com/41143

Answer 3

thank you! will doo hahaha at this point honestly anything helps

Answer 4

You need a solid footing in the world of factorials... the numbers that like to shout :D

Answer 5

You know what \(\Large n!\) means?

Answer 6

Yes I doo!!

Answer 7

@uri

Answer 8

Anyway.... if we have ten different, say, marbles instead, and by different, I mean no pairs... each marble unique

How many ways can we arrange ten of them in a row?

Answer 9

crossing my fingers that it's
10!
= 10x9x8x7x6x5x4x3x2x1
= 3628800 ways

Answer 10

That's right :D
10!

In fact, if we had ten socks (which are different) instead, then we would have 10! arrangements

Answer 11

The problem here is that we have pairs.... five of them, which are exactly the same... can you see why that's the problem? :3

Answer 12

I KNOW it's basically killing mee ahhh!!!

Answer 13

Yeah, relax.... I'm here, aren't I? :3

Answer 14

Anyway... here's a scenario...
\[\Large \color{red}l\quad\color{green}l\quad\color{blue}l\quad\color{red}l\quad\color{violet}l\quad\color{orange}l\quad\color{orange}l\quad\color{violet}l\quad\color{green}l\quad\color{blue}l\]

Answer 15

this makes sense so far?

Answer 16

yes!

Answer 17

Now, the problem with blindly considering 10! outright, despite having 5 identical pairs is...

Answer 18

10! also counts the "other" arrangement, where, say, you switch those two red socks... but that'd give the exact same arrangement, right?

Answer 19

yep!!

Answer 20

So in 10!, exactly half of those just switches the pair of red socks... right?

Answer 21

yep! would that be a permutation inside of a factorial??

Answer 22

OH I GOT IT!!!! 10!/(2p2)^5)

Answer 23

thank youuu!!!!

Answer 24

Strictly speaking it's
\[\Huge \frac{_{10}P_{10}}{(_2P_2)^5}\]
But then again, it's easy to show that
\[\Large _nP_n = n!\]

Answer 25

Any idea why 2P2 was raised to 5?

Answer 26

there are five pairs so if you multiply 2P2 five times it'd go to the fifth power! thank you again!!!

Answer 27

:)