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**confusingmath** When rolling three dice, what is the probability of

a) getting three different numbers?

Mr F says: The total number of permutations is 6^3. The number of these with all numbers different is (6)(5)(4) = ..... Therefore .....

b) having at least one two and at least one four?

Mr F says: Consider

2, 4, x

4, 2, x

2, x, 4

4, x, 2

x, 2, 4

x, 4, 2

(or you could just arrange 2, 4, x in 3! ways)

where x can be any number from 1 to 6. So there are (4)(6) = 24 ways.

c) getting two threes given that no more than one five was rolled?

Mr F says: Consider

x, 3, 3 can be arranged in 3! = 6 ways and there are 5 values that x can have (x cannot equal 3) ....

Any helpful hints will be greatly appreciated.