How many ways can you arrange 3 French books, 2 history books, and 5 math books if (a) all books are different? (b) all French books look the same, all history books look the same, and all math books look the same?

Question

How many ways can you arrange 3 French books, 2 history books, and 5 math books if (a) all books are different? (b) all French books look the same, all history books look the same, and all math books look the same?

Asked on March 26, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-03-26T13:44:43.786Z

Here are the solutions for Exercise 2: Exercise 2: How many ways can you arrange 3 French books, 2 history books, and 5 maths books? Total number of books = 3 (French) + 2 (History) + 5 (Maths) = 10 books. a) All books are different Step 1: Determine the total number of books. $$N = 3 + 2 + 5 = 10$$ Step 2: If all books are different, then arranging them is a permutation of $$N$$ distinct items. The number of ways is $$N!$$. $$ \text{Number of ways} = 10! $$ Step 3: Calculate the factorial. $$ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800 $$ The number of ways to arrange the books if all books are different is $$\boxed{3,628,800}$$. b) All the French books look the same, all the history books look the same and all the maths books look the same. Step 1: Identify the total number of books ($$N$$) and the number of identical books in each category. $$N = 10$$ Number of French books ($$n_F$$) = 3 Number of History books ($$n_H$$) = 2 Number of Math books ($$n_M$$) = 5 Step 2: Use the formula for permutations with repetitions: $$ \text{Number of ways} = \frac{N!}{n_F! n_H! n_M!} $$ Step 3: Substitute the values and calculate. $$ \text{Number of ways} = \frac{10!}{3! 2! 5!} $$ $$ \text{Number of ways} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{ (3 \times 2 \times 1) \times (2 \times 1) \times 5!} $$ $$ \text{Number of ways} = \frac{10 \times 9 \times 8 \times 7 \times 6}{3 \times 2 \times 1 \times 2 \times 1} $$ $$ \text{Number of ways} = \frac{30240}{12} $$ $$ \text{Number of ways} = 2520 $$ The number of ways to arrange the books with identical books in each category is $$\boxed{2520}$$.

Accepted Answer · 2026-03-26T13:44:43.786Z

Here are the solutions for Exercise 2: Exercise 2: How many ways can you arrange 3 French books, 2 history books, and 5 maths books? Total number of books = 3 (French) + 2 (History) + 5 (Maths) = 10 books. a) All books are different Step 1: Determine the total number of books. $$N = 3 + 2 + 5 = 10$$ Step 2: If all books are different, then arranging them is a permutation of $$N$$ distinct items. The number of ways is $$N!$$. $$ \text{Number of ways} = 10! $$ Step 3: Calculate the factorial. $$ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800 $$ The number of ways to arrange the books if all books are different is $$\boxed{3,628,800}$$. b) All the French books look the same, all the history books look the same and all the maths books look the same. Step 1: Identify the total number of books ($$N$$) and the number of identical books in each category. $$N = 10$$ Number of French books ($$n_F$$) = 3 Number of History books ($$n_H$$) = 2 Number of Math books ($$n_M$$) = 5 Step 2: Use the formula for permutations with repetitions: $$ \text{Number of ways} = \frac{N!}{n_F! n_H! n_M!} $$ Step 3: Substitute the values and calculate. $$ \text{Number of ways} = \frac{10!}{3! 2! 5!} $$ $$ \text{Number of ways} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{ (3 \times 2 \times 1) \times (2 \times 1) \times 5!} $$ $$ \text{Number of ways} = \frac{10 \times 9 \times 8 \times 7 \times 6}{3 \times 2 \times 1 \times 2 \times 1} $$ $$ \text{Number of ways} = \frac{30240}{12} $$ $$ \text{Number of ways} = 2520 $$ The number of ways to arrange the books with identical books in each category is $$\boxed{2520}$$.