Find the number of ways in which 22 different books can be given to 5 students, so that two students get 5 books each and all the remaining students get 4 books each.

Explanation

To solve the problem of distributing 22 different books to 5 students such that two students receive 5 books each and the remaining three students receive 4 books each, we can follow these steps:

### Step 1: Identify the distribution of books
We have:
- 2 students (let's say A and B) who will receive 5 books each.
- 3 students (let's say C, D, and E) who will receive 4 books each.

### Step 2: Calculate the total number of books distributed
The total number of books distributed is:
- For students A and B: \(5 + 5 = 10\)
- For students C, D, and E: \(4 + 4 + 4 = 12\)

Thus, the total number of books is \(10 + 12 = 22\), which matches the total number of books we have.

### Step 3: Choose the students who will receive 5 books
We need to choose 2 students from the 5 to receive 5 books each. The number of ways to choose 2 students from 5 is given by the combination formula:
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

### Step 4: Distribute the books to the chosen students
Next, we need to distribute the books:
- Choose 5 books for student A from 22 books: This can be done in \(\binom{22}{5}\) ways.
- Choose 5 books for student B from the remaining 17 books: This can be done in \(\binom{17}{5}\) ways.
- The remaining 12 books will be distributed among students C, D, and E, with each receiving 4 books.

### Step 5: Distribute the remaining books
Now we need to distribute the remaining 12 books:
- Choose 4 books for student C from 12 books: This can be done in \(\binom{12}{4}\) ways.
- Choose 4 books for student D from the remaining 8 books: This can be done in \(\binom{8}{4}\) ways.
- The last 4 books automatically go to student E.

### Step 6: Calculate the total arrangements
Now, we need to consider the arrangements of the students who received the books. Since A and B are indistinguishable in terms of receiving 5 books, we will divide by \(2!\) (for A and B). Similarly, since C, D, and E are indistinguishable in terms of receiving 4 books each, we will divide by \(3!\).

### Step 7: Combine all the calculations
Putting it all together, the total number of ways to distribute the books is:
\[ \text{Total ways} = \binom{5}{2} \times \binom{22}{5} \times \binom{17}{5} \times \binom{12}{4} \times \binom{8}{4} \times \frac{1}{2!} \times \frac{1}{3!} \]

### Step 8: Simplify the expression
Now we can simplify the expression to find the final answer.

### Final Calculation
1. Calculate each binomial coefficient:
- \(\binom{5}{2} = 10\)
- \(\binom{22}{5} = \frac{22!}{5!(22-5)!}\)
- \(\binom{17}{5} = \frac{17!}{5!(17-5)!}\)
- \(\binom{12}{4} = \frac{12!}{4!(12-4)!}\)
- \(\binom{8}{4} = \frac{8!}{4!(8-4)!}\)

2. Combine the results:
\[ \text{Total ways} = 10 \times \binom{22}{5} \times \binom{17}{5} \times \binom{12}{4} \times \binom{8}{4} \times \frac{1}{2} \times \frac{1}{6} \]

### Final Answer
The final answer will be a numerical value obtained from the above calculations.

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