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Mathematics 8 Online
OpenStudy (anonymous):

In how many ways can you arrange 6 CD’s from a collection of 50 CD’s?

OpenStudy (usukidoll):

8 piles at 6 CDs a piece would be 48 CDs with 2 CDS left over

OpenStudy (anonymous):

what O.o

OpenStudy (usukidoll):

you're talking about Compact Discs right? Not the CD from the bank right?

OpenStudy (anonymous):

i dunno. the question thingy said CD's , however i don't understand why what CD stands for would matter.

Directrix (directrix):

@anthony~ In this question, it sounds as if six CDs will be chosen and arranged on a shelf. So, the order counts. And, this selection of six CDs will be done exactly one time. There are 50 choices for the first CD, 49 for the second, 48 for the third, and as the pattern continues, 45 choices for the sixth CD. By the Multiplication Rule for Counting, the number of arrangements is 50 * 49 * __ * __ * __ * __ = ?

OpenStudy (anonymous):

i got : 1,144,130,000

OpenStudy (usukidoll):

O______________O

OpenStudy (anonymous):

stupid math

Directrix (directrix):

@anthony~ I got the same answer as yours. http://www.wolframalpha.com/input/?i=50*49*48*47*46*45%3D

OpenStudy (usukidoll):

at least you're not writing proofs. Look at my question. Up for two days yo

OpenStudy (kropot72):

So the correct answer is 11, 441, 304, 000

OpenStudy (anonymous):

thats a lot of combinations

OpenStudy (anonymous):

50C6..

OpenStudy (shubhamsrg):

P(50,6) yep, thats right.

OpenStudy (anonymous):

Here arrangement refers to Permutations . So here we need to Arrage 6 CD'S from a pile of 50 CD'S Permutations ( P ) = n p r P = \[\frac{ n! }{ (n-r)! }\] So here n = 50 , r= 6 P = \[\frac{ 50! }{ 44! }\]

OpenStudy (anonymous):

The word "arrange" in your question is open to interpretation. From the 50 CD's, you can draw #1 from 50, #2 from 49, #3 from 48, #4 from 47, #5 from 46, and #6 from 45. The number of ordered sets of 6 CD's that you could have drawn is 50 * 49 * 48 * 47 * 46 * 45, or 50!/(50-6)! These are ordered sets; that is, set ABCDEF (where each letter represents a unique CD), and set BACDEF were both counted, with every other permutation of these six letters. If you just want to know how many different sets, order not important, you have to divide by the number of ways you can arrange 6 items. That is 6!. So that answer would be 50!/[(6!)(50-6)!]

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