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catgirl17
 3 years ago
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?
catgirl17
 3 years ago
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?

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Ledah
 3 years ago
Best ResponseYou've already chosen the best response.0Seeing 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

mathplane
 3 years ago
Best ResponseYou've already chosen the best response.1This 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 1469...)... So, your final solutions is (9 x 8 x 7 x 6)/(4!) = 126

mathplane
 3 years ago
Best ResponseYou've already chosen the best response.1Using the formula for combinations: 9C4 = 9!/5!4! = 126

Ledah
 3 years ago
Best ResponseYou've already chosen the best response.0Totally thought of something else.......

catgirl17
 3 years ago
Best ResponseYou've already chosen the best response.0cause when you do 9x8x7x6=3024 and then if you divide it by 4 you get 756

mathplane
 3 years ago
Best ResponseYou've already chosen the best response.1No, you must divide by 4! 3024/24

catgirl17
 3 years ago
Best ResponseYou've already chosen the best response.0where did you get 24 from?

mathplane
 3 years ago
Best ResponseYou've already chosen the best response.14! = 4 x 3 x 2 x 1 = 24

jsoeung
 3 years ago
Best ResponseYou've already chosen the best response.0The exclamation mark isn't yelling, it's called a factorial :) so N! would be N multiplied by all the numbers that come before it
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