Final answer:
There are 150 different ways to distribute 5 balls into 3 distinct urns, ensuring no urn is empty, by using the principle of inclusion-exclusion on the initial 3^5 total distributions and adjusting for overcounting.
Step-by-step explanation:
The problem is to calculate in how many ways can 5 balls, each of a different color, be distributed among 3 distinct urns such that no urn remains empty. This is a combinatorial problem that involves assigning the balls to the urns with the constraint that each urn must contain at least one ball.
To solve this, we can use the principle of inclusion-exclusion. There are 3 ways to assign each ball to an urn, so initially, we have 3^5 total distributions. However, this count includes the cases where one or more urns are empty.
To adjust for this, we subtract the distributions where at least one urn is empty. There are 3 choices for which urn is empty and 2^5 ways to distribute the balls to the remaining two urns, leading to 3 × 2^5 to be subtracted. But now we've subtracted too much, as we've removed the cases with two empty urns twice. Since there's exactly one way to put all balls into one urn, we add back the 3 distributions where all balls are in a single urn.
So the total number of distributions without any urns being empty is 3^5 - 3 × 2^5 + 3. Calculating this gives us 243 - 96 + 3 = 150 different ways.