Dice - thrown 3 times & equal sides

[Brhum]

A dice is thrown three times .
Calculate the ways to get at least two sides are equal
soluation
90+6=96
or
30+6=36

[Guest]

The question is whether "permutations" count as different solutions or whether we are only interested in "combinations".
I.e. is rolling 1,1,2 different from 1,2,1?

If we are only interested in combinations (by which I mean the order doesn't matter), then there are 6 ways of rolling the same number all three times (e.g. 1,1,1 or 2,2,2 etc.), and there are 6x5 = 30 ways of rolling so that exactly two rolls are the same (e.g. 1,1,2 or 6,6,4) because we have 6 choices for the number that appears twice and 5 choices for the number that appears once (there is one less choice because we have to avoid the same number appearing 3 times). In total that makes 36 ways.

If we are interested in permutations as well (by which I mean the order does matter), then again there are 6 ways of rolling the same number all three times, and there are 3x30 = 90 ways of rolling so that exactly two rolls are the same (the number that appears only once can appear in one of three places at the start, middle, or last roll, meaning there are 3 times as many ways then there were for the "combinations" case). In total this makes 96 ways.

Both answers are correct because the question is ambiguous, but if you have to choose one I would go with 96, as usually you are more interested in permutations. With dice rolls you are often interested in probabilities and if you were asked "what is the probability that at least two rolls are the same?" the answer is 96/(6x6x6) which is a lot easier to calculate from the permutations than from the combinations.

Hope this helped,

R. Baber.

[Brhum]

thanks ,Guest

[leesajohnson]

30+6 = 36 it the right answer for the question you have asked.