Let us solve this step by step:

Problem:

Emon has 10 different books (5 of which he really likes). He has 7 friends and decides to:

1. Give one book to each of his 7 friends.
2. Keep 3 books for himself (out of his favorite books).

We need to determine the total number of different ways he can do this.

Step 1: Choose the 3 books Emon keeps

Since Emon wants to keep 3 of his favorite books and there are 5 favorite books, the number of ways to choose 3 books from 5 is calculated using combinations:

$\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \cdot 4}{2 \cdot 1} = 10$

So, there are 10 ways for Emon to choose the 3 favorite books he keeps.

Step 2: Assign the remaining 7 books to his 7 friends

After Emon keeps 3 books, there are $10 - 3 = 7$ books left. Each of these 7 books is given to one friend, and since all books are different, the arrangement of books among the 7 friends matters. This is a permutation problem.

The number of ways to assign 7 books to 7 friends is:

$7! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5040$

Step 3: Total number of ways

To find the total number of ways Emon can distribute the books, we multiply the results of Step 1 and Step 2:

$\text{Total ways} = \binom{5}{3} \cdot 7! = 10 \cdot 5040 = 50400$

Final Answer:

Emon can distribute the books in 50,400 different ways.

Do you want further clarification or additional details?
Here are 5 related questions to deepen your understanding:

1. How would the solution change if all 10 books were equally liked by Emon?
2. What if Emon wanted to keep 4 books instead of 3?
3. How is the calculation affected if Emon has fewer than 7 friends?
4. Can you explain why combinations were used for selecting books for Emon and permutations for assigning them to friends?
5. How would this problem change if some books were identical instead of unique?

Tip: Remember that combinations are used when the order doesn't matter, while permutations are used when the order matters!