anonymous
  • anonymous
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 :)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Thanks so much for the reply!! I forgot to add that the answer is 113, 400 :)
Dean.Shyy
  • Dean.Shyy
Maybe, if you check out topics on combinations and permutations, it would help. Take a look at this: http://url.moosaico.com/41143
anonymous
  • anonymous
thank you! will doo hahaha at this point honestly anything helps

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

terenzreignz
  • terenzreignz
You need a solid footing in the world of factorials... the numbers that like to shout :D
terenzreignz
  • terenzreignz
You know what \(\Large n!\) means?
anonymous
  • anonymous
Yes I doo!!
anonymous
  • anonymous
@uri
terenzreignz
  • terenzreignz
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?
anonymous
  • anonymous
crossing my fingers that it's 10! = 10x9x8x7x6x5x4x3x2x1 = 3628800 ways
terenzreignz
  • terenzreignz
That's right :D 10! In fact, if we had ten socks (which are different) instead, then we would have 10! arrangements
terenzreignz
  • terenzreignz
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
anonymous
  • anonymous
I KNOW it's basically killing mee ahhh!!!
terenzreignz
  • terenzreignz
Yeah, relax.... I'm here, aren't I? :3
terenzreignz
  • terenzreignz
Anyway... here's a scenario... \[\Large \color{red}l\quad\color{green}l\quad\color{blue}l\quad\color{red}l\quad\color{violet}l\quad\color{orange}l\quad\color{orange}l\quad\color{violet}l\quad\color{green}l\quad\color{blue}l\]
terenzreignz
  • terenzreignz
this makes sense so far?
anonymous
  • anonymous
yes!
terenzreignz
  • terenzreignz
Now, the problem with blindly considering 10! outright, despite having 5 identical pairs is...
terenzreignz
  • terenzreignz
10! also counts the "other" arrangement, where, say, you switch those two red socks... but that'd give the exact same arrangement, right?
anonymous
  • anonymous
yep!!
terenzreignz
  • terenzreignz
So in 10!, exactly half of those just switches the pair of red socks... right?
anonymous
  • anonymous
yep! would that be a permutation inside of a factorial??
anonymous
  • anonymous
OH I GOT IT!!!! 10!/(2p2)^5)
anonymous
  • anonymous
thank youuu!!!!
terenzreignz
  • terenzreignz
Strictly speaking it's \[\Huge \frac{_{10}P_{10}}{(_2P_2)^5}\] But then again, it's easy to show that \[\Large _nP_n = n!\]
terenzreignz
  • terenzreignz
Any idea why 2P2 was raised to 5?
anonymous
  • anonymous
there are five pairs so if you multiply 2P2 five times it'd go to the fifth power! thank you again!!!
terenzreignz
  • terenzreignz
:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.