Two different answers - cubes and colors If there are 5 cubes in 5 different colors (on each cube the numbers 1-6), and I want all the ways to choose the cubes so that at least 1 cube shows the number '3'. I can think of two different ways to do this, but for some reason they bring different answers!One way is to subtract all the ways of getting no 3's from all the ways of getting anything, and then multiply by all the ways for the different colors $=(6^{5}-5^{5})\cdot5!$
Second way is to take 1 cube and give it the number 3, and then multiply by all the other results and then multiply by all the ways for the different colors $=(1\cdot6^{4})\cdot5!$So why are these two answers different?
 A: Your first solution (i.e. $(6^5 - 5^5)5!$) is correct.
In the second solution you are picking a particular cube (let's say the red cube) and constraining it to show 3. But there are valid cases where a different cube (say the blue cube) is showing 3 and the red cube is not showing 3.
A: Both of your approaches are slightly wrong, though they are sound in concept.
In the first attempt, you reasoned correctly that the difference between the total number of combinations ($6^5$) and the number of combinations without choosing the face $3$ in any dice ($5^5$). We then can count the ways this can be permutated multiplying the result by 5!.
The second way you try to do it is also valid.  You would need to choose one color to pick the 3 face on it (5 ways to do this) and then choose the faces combinations in the remaining 4 dices, for a total of $5\times 6^4$ combinations.
BUT in this second attempt we have made a mistake, and that is counting some combinations more than one. To see why this is the case, think what would happen if we had two dices instead of five. We could set the first dice to three, and there would be 6 ways to set the second dice. Or we could set the second dice to 3, and there would be 6 ways to pick the face of the fist one. However, one of those combinations is actually the same: picking 3 in both dices.
Therefore, the first attempt was correct, but not the second. Hope this clarifies the mistake.
