Question:
Three dice (six faces: each face -> number 1 to 6) are rolled. What is the number of possible outcomes such that at least one die shows number 2?
My attempt:
One die has to show two. The possible ways in which the other two can show their numbers is $6 \cdot 6 = 36$. But, these dices can be interchanged in $3$ ways. Thus, answer $=36 \cdot 3 = 108$.
But it is wrong. I need help identifying where I went wrong.
On
You are double counting the outcomes. You have to assume there is exactly one 2 instead of at least one 2, then exactly two 2's, and finally exactly three 2's:
Exactly one 2: $1\times 5\times 5\times \binom{3}{1}=75$
Exactly two 2's: $1\times 1\times 5\times \binom{3}2=15$
Exactly three 2's: $1\times 1\times1 =1$
This gives $75+15+1=91$ desirable outcomes.
An easier method would be to just find the number of outcomes with no 2's, then subtract that from the total number of outcomes:
$$6\cdot 6\cdot 6 - 5\cdot 5\cdot 5=216-125=91$$
You are counting twice some of the outcomes. When you said that you fix one dice in "2" and you count all possible outcomes of the two remaining dices, what happen if you get another "2"?
As an example consider the following configuration: when you first fix "2" as the first outcome, you get $\color{blue}{2},2,3$. Now, when you moved the fixed "2" to the second outcome you get again $2,\color{blue}{2},3$.
Hint if you need that at least 1 dice shows "2", you have to take out from all the possibilities the outcomes that no dice shows "2".