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Combinations - Help!
There are seven dice. Each die has six faces. How many different combinations are there of these seven dice?
Please explain how you use the Combination / Permutation Formulas to arrive at the answer.
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I'm not 100% sure what the Q is looking for, but if it counts 1222222 and 21222222 as two different arrangements (hence ordered) then its:
otherwise, wouldn't you have to build for every condition of the combination. and then add them i.e.
all 7 the same: 6 diff combs * (7!/7!)
6 same + 1 diff = (7!/(1!*6!)) * 6
5 same + 2 same = 7!/(2!*5!) * 6
and then for all of the different permutations? That looks like a beast. Any comments
Originally Posted by mac999993
We need to partition the numbers from 1-6 to get a 7 digit sequence where a number can repeat upto 7 times.
6 distinct numbers => 5 partitions.
Lets see how this might work for a given 7-digit sequence -
2233566 can be written as |22|33||5|66 where a bar (|) denotes a partition and the empty space between 2 bars or to the left of the 1st bar denotes a missing number (in the above example, 1 and 4).
So in totality we have 7 digits and 5 bars . i.e., we have 12 positions that the 5 bars can take.
12C5 = 792
And that is how many 7 dice combos you can have.
Note: This question seems to be a little above GMAT level!
Last edited by arjmen; 02-07-2006 at 04:16 AM.
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