Let's solve the problem step-by-step!

(a) How many different patterns are there in total?

We are distributing $7$ identical balls into $5$ distinct boxes. This is a stars and bars problem where the number of ways to do this is given by: $\binom{n + k - 1}{k - 1} = \binom{7 + 5 - 1}{5 - 1} = \binom{11}{4}.$ $\binom{11}{4} = \frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1} = 330.$

So, there are 330 different patterns in total.

(b) How many patterns are there such that exactly 4 boxes are empty?

If 4 boxes are empty, this means all 7 balls must go into 1 box (since the boxes are distinct). There are $\binom{5}{1} = 5$ ways to select which box is non-empty.

Thus, there are 5 patterns.

(c) How many patterns are there such that exactly 2 boxes are empty?

If 2 boxes are empty, 7 balls are distributed among 3 boxes. We must calculate the number of patterns using partitions of $7$ into at most $3$ parts such that no box is empty. This corresponds to the solutions of: $x_1 + x_2 + x_3 = 7 \quad \text{where } x_i \geq 1.$

Let $x'_i = x_i - 1$. Substituting, we solve: $x'_1 + x'_2 + x'_3 = 4, \quad x'_i \geq 0.$

The number of non-negative integer solutions is: $\binom{4 + 3 - 1}{3 - 1} = \binom{6}{2} = 15.$

Now, we select which 3 boxes are used (out of 5): $\binom{5}{3} = 10.$

Thus, the total number of patterns is: $15 \cdot 10 = 150.$

There are 150 patterns.

(d) What is the probability that exactly 2 boxes are empty?

From part (a), the total number of patterns is $330$, and from part (c), the number of patterns with exactly 2 empty boxes is $150$. The probability is: $P = \frac{150}{330} = \frac{15}{33} = \frac{5}{11}.$

The probability is $\frac{5}{11}$.

Summary of Answers:

(a) $330$
(b) $5$
(c) $150$
(d) $\frac{5}{11}$

Do you want detailed explanations of any specific step?
Here are 5 related questions to deepen understanding:

1. How does the stars and bars theorem work in combinatorics?
2. Can you explain why partitions are used in part (c)?
3. How would the problem change if the balls were distinguishable?
4. What if there were additional constraints (e.g., no box could have more than 5 balls)?
5. How would the probability change if the total number of balls increased to 10?

Tip: Always verify combinatorial formulas (like $\binom{n}{k}$) for correctness when solving step-by-step!