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Futuristic's gambling laboratory has decided that 6-sided

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Futuristic's gambling laboratory has decided that 6-sided [#permalink] New post 09 Sep 2006, 15:44
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Futuristic's gambling laboratory has decided that 6-sided dice are out of fashion, and creates dice that are 10 sided, with numbers from 1 to 10 on the faces.

How many different combinations of numbers can be found by throwing 3 of these dice together? Assume all dice are undistinguishable.

E.g.

1,9,8 (is the same as 9,8,1)
1,9,9
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 [#permalink] New post 09 Sep 2006, 16:10
It's 720

the first dice can face any fo the 10 digits.
The second dice can face 9 digits, leaving the 1 digit occurred in dice 1
The third dice can face 8 digits, leaving the 2 digits occurred in dice 1 and 2

SO v have a a totoal number of 10*9*8 number of digits that V can have = 720
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 [#permalink] New post 09 Sep 2006, 16:30
nope...the question does not say that they are all different values
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 [#permalink] New post 09 Sep 2006, 16:36
I think 120

1,9,8 is the same as 1,8,9 and 9,8,1

Hence answer = 10C3 = 10*9*8/3*2*1 = 120
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 [#permalink] New post 09 Sep 2006, 17:03
Mmmm, thinking again I’m now gettin 340

10^3-(10P3)/2-300=340 :?: :?:
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 [#permalink] New post 09 Sep 2006, 18:15
gmatornot wrote:
I think 120

1,9,8 is the same as 1,8,9 and 9,8,1

Hence answer = 10C3 = 10*9*8/3*2*1 = 120


This is correct only if we assume all dice have different values. What if they all have same values....or 2 of 3 have the same values?

hint hint....
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 [#permalink] New post 09 Sep 2006, 19:10
Ok, plz omit my previous posts.

If in the following e.g

1,9,3 and 1,3,9 and 3,1,9 and 3,9,1 and 9,1,3 and 9,3,1 --> counts as 1
9,9,1 and 1,9,9 and 9,1,9 --> counts as 1
9,9,9 --> counts as 1

Then it´s
10C3 + (10*10) + 10 = 230 :?: :roll:
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 [#permalink] New post 09 Sep 2006, 19:38
you're getting closer.....

of the cases in (10x10), are there any also falling into 3rd part (10)....
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 [#permalink] New post 11 Sep 2006, 01:49
There are three cases:

(1) All three show the same number - 10 cases
(2) Two dice are the same, one is different 10*9
(3) All three are different 10C3=10*9*8/(3*2)=10*12


Total 10*(1+9+12)=220

Last edited by kevincan on 11 Sep 2006, 12:45, edited 1 time in total.
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 [#permalink] New post 11 Sep 2006, 10:35
kevincan wrote:
There are three cases:

(1) All three show the same number - 10 cases
(2) Two dice are the same, one is different 10*9
(3) All three are different 10*9*8


Total 10*(1+9+72)=820


(3) All three are different 10*9*8


In the above calculation, why are you finding the number of permutations? The dice are identical.
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 [#permalink] New post 11 Sep 2006, 12:35
It would be 3^10. Using the following logic.

For 2 dice with 6 sides it is 2^6. So for 3 dice with 6 faces it is 3^6.Hence for 3 dice with 10 faces it is 3^10.
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 [#permalink] New post 11 Sep 2006, 12:46
Futuristic wrote:
kevincan wrote:
There are three cases:

(1) All three show the same number - 10 cases
(2) Two dice are the same, one is different 10*9
(3) All three are different 10*9*8


Total 10*(1+9+72)=820


(3) All three are different 10*9*8


In the above calculation, why are you finding the number of permutations? The dice are identical.


You're right- I've corrected my answer! My holiday was too long
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 [#permalink] New post 11 Sep 2006, 12:48
Again, kevin shows us the light. The OA is 220.
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 [#permalink] New post 11 Sep 2006, 13:04
Why is this not

10x10x10/3!

What are we overcounting in this case?

Thanks/
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 [#permalink] New post 11 Sep 2006, 13:19
Futuristic wrote:
Again, kevin shows us the light. The OA is 220.


Sorry about my mistake! Nice question- where did you get it-did you make it up?
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 [#permalink] New post 11 Sep 2006, 13:47
yes kev, i made it up. i found another question similar to this with 3 regular dice....and just adapted it a little to make it more interesting.

for those that are interested, with 3 regular dice there are 56 possibilities in all....
  [#permalink] 11 Sep 2006, 13:47
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