anonymous
  • anonymous
you own 9 pairs of jeans and are taking 4 of them on vacation. in how many ways can you choose 4 pairs of jeans from the 9?
Mathematics
jamiebookeater
  • jamiebookeater
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Ledah
  • Ledah
Seeing as this is probability I'm going to say that there are 36 or 45 choices on pants you can pick. hope this helps.. there are so many ways you can do this
anonymous
  • anonymous
9c4
anonymous
  • anonymous
This is an example of a combination problem... First, establish how many ways there are to pack the jeans.... There are 9 choices for the 1st pair of pants... 8 remaining choices for the 2nd pair... 7 remaining for the 3rd pair... And, 6 remaining choices for the 4th pair.... this implies 9 x 8 x 7 x 6 permutations.... However, ORDER DOES NOT MATTER.. Therefore, we must divide to get rid of all the "double counting"... (i.e. choosing 1469 pants is the same as 4169 or 9146... in each case, you packed 1-4-6-9...)... So, your final solutions is (9 x 8 x 7 x 6)/(4!) = 126

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anonymous
  • anonymous
Using the formula for combinations: 9C4 = 9!/5!4! = 126
Ledah
  • Ledah
Totally thought of something else.......
anonymous
  • anonymous
how did you get 126?
Ledah
  • Ledah
Thats what I'm wondering
anonymous
  • anonymous
cause when you do 9x8x7x6=3024 and then if you divide it by 4 you get 756
anonymous
  • anonymous
No, you must divide by 4! 3024/24
anonymous
  • anonymous
where did you get 24 from?
anonymous
  • anonymous
4! = 4 x 3 x 2 x 1 = 24
anonymous
  • anonymous
The exclamation mark isn't yelling, it's called a factorial :) so N! would be N multiplied by all the numbers that come before it
anonymous
  • anonymous
Just fyi

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