SOLUTION: hey i'm having trouble on this question i've tried 7 times and different equations but i just cant get it: There are 24 students in your class. Your teacher wants to put you in

Question 1052575: hey i'm having trouble on this question i've tried 7 times and different equations but i just cant get it:
There are 24 students in your class. Your teacher wants to put you in groups for your data-
management project. How many ways can six groups of four be made if
A)there are no restrictions
Answer=4.509 x 10^12 (will also accept 3.247 x 10^15)
Thankyou.:)

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
i think i found a formulas you are looking for.

i tried:

c(24,4) * c(20,4) * c(16,4) * c(12,4) * c(8,4) * c(4,4)

the answer i got was 3.246670537 * 10^15.

round that to 3 decimal digits and you get 3.247 * 10^15.

the rationale behind it is as follows:

out of 24 people, you can make 24c4 groups of 4.

once you select 4 of those, there are 20c4 groups of 4 that can be made out of the rest of them.

once you select 4 of those, there are 16c4 groups of 4 that can be made out of the rest of them.

and so on.

the combination formula used is c(n,x) which can also be shown as ncx.

this formula is n! / (x! * (n-x)!)

for example: c(24,4) = 24c4 = 24! / (4! * 20!)

i then divided the answer by 6! and i got 4.509 * 10^12.

i confess i didn't really know if these formulas would be right, but made some semi-educated guesses that turned out to be right.

so your answer is either:

c(24,4) * c(20,4) * c(16,4) * c(12,4) * c(8,4) * c(4,4)

or:

c(24,4) * c(20,4) * c(16,4) * c(12,4) * c(8,4) * c(4,4) / 6!.

the division by 6! has to do with taking away ordering of the sets of 4.

what you wind up with is multiples sets of 6 groups of 4 with the same elements in each group only in a different arrangement.

the division by 6! takes away the different arrangements.

it's har to explain and even harder to show, but my suspicions were correct and i did come up with the right answer based on what answers were acceptable.

i'm happy just to be able to do that.

what makes these problems hard is not being able to verify that your formula is correct without going through hoops because the number of possibilities is so large.

i usually try to reduce the number of possibilities and go through the manual labor of identifying each possibility.

that was hard to do in this case, but i was able to do enough to develop my hunches.

hope this satisfies you.

sorry i can't explain why i did what i did any better.

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