You have three Die - what are the possibilities of getting the same number on two of them. I thought the answer would be 6.6.5= 180 the first two throws would be 6 because those numbers would still be available but in the last throw the number would be 5 to reduce the numbers still left. The answer is 90 - how is my answer double this amount.
Im new at combinatorics so your help would be appreciated.
On
The number of ways of selecting 2 out of 3 dice (which show identical numbers) is $^3C_2 = 3$. Both of them can show either of $\{1, 2, \ldots, 6\}$, so $6$ possibilities. Finally, the remaining die can show any of the remaining $5$ numbers. The total number of ways the whole scenario can play out is therefore $3\times6\times 5 = 90$.
I would suggest you to subtract the possibilities, where we have different numbers on each die and we have the same number on all of the dices, from all of the possibilities. In this case, let $A_1$ be the event where all of the dice have different values and $A_2$ be the event where all the dice have the same value. Then $|A_1| = 6 \cdot 5 \cdot 4 = 120$ and $|A_2| = 6$ so the answer becomes $$6^3 - 6 \cdot 5 \cdot 4 - 6 = 90$$