anonymous
  • anonymous
how many possible combinations can there be with the numbers 6,6,4,4,1?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
if they were all different you would have 6! combinations, but since you cannot tell the 6s apart, nor the twos, it is \[\frac{6!}{2\times2}\]
anonymous
  • anonymous
"6,6,4,4,1" are 5 numbers isn't ?
anonymous
  • anonymous
lol

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anonymous
  • anonymous
yeah i guess it is isn't it!
anonymous
  • anonymous
so if it was 42333 what would it be? whats the general formula?
anonymous
  • anonymous
what is an additional number between friends? ok i was wrong, maybe try \[\frac{5!}{2\times 2}\] if you want to actaully get the correct answer
anonymous
  • anonymous
haha :D
anonymous
  • anonymous
\[\frac{5\times 4\times 3\times 2}{2\times 2}=5\times 3\times 2=30\] (did i get that right?)
anonymous
  • anonymous
looks correct, but what is the general formula?
anonymous
  • anonymous
\[\frac{5!}{3!}\] for the second one
anonymous
  • anonymous
so that otheer one is really 2!x2! in the denominator?
anonymous
  • anonymous
yes, if you want to think of it that way. the number of ways you can permute 2 things is 2! = 2
anonymous
  • anonymous
awesome thanks so much!
anonymous
  • anonymous
yw
anonymous
  • anonymous
sorry about mis-counting
anonymous
  • anonymous
its all good... all worked out!

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