# When six dice are rolled, the number of different numbers which can appear can range from 1 to 6. In how many different ways can you have exactly 4 different numbers?

We want the number of different ways there are of rolling exactly four different dice. This is the same as the number of ways of rolling exactly one pair.

There are six possible pairs we could roll and "6 choose 2" 6!/(2!4!) = 15 ways this pair could be arranged amongst the six dice.

Conditional on a pair being rolled, there are five possible numbers the four remaining dice can be, but they must be different to each other. There are "5 choose four" 5!/(4!1!) = 5 ways of choosing the four different numbers and 4! ways those four numbers can be arranged amongst the four remaining dice.

Multiplying all these possible ways together we get

Number of ways = 6*15*5*4! = 10800 (the probability is 10800/(6^6)= 0.231)

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There are six different dice (they don't have labels, and they have precisely the same properties, but they are different dice).

We want the probability of exactly two having the same value and the other four having different values to those two and to each other. Because these are 'and' requirements on the dice, we multiply the probabilities together and then multiply by the number of possible configurations of the dice.

Now, the probability of two of the dice having the same value is (1/6)^2 = 1/36

The probabilities (in order) of the other four dice having different values to

this value and to each other are 5/6, 4/6, 3/6, 2/6

There are 6!/(2!4!) ways of choosing the two dice from the six and 4!

possible combinations of the remaining 4 (4! ways).

Combining this all together, the probability P of exactly two dice having the

same value (the other four must have different values) is

P = 6!/(2!4!) (1/36) 4! (5/6)(4/6)(3/6)(2/6) = 0.925926

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