The number of ways of distributing `3` identical physics books and `3` identical mathematics books among three students such that each student gets at least one books is

Options

A. `45`
B. `55`
C. `64`
D. `72`

Explanation

To solve the problem of distributing 3 identical physics books and 3 identical mathematics books among three students such that each student gets at least one book, we can follow these steps:

### Step 1: Calculate the total ways to distribute the books without restrictions.
We can use the stars and bars method for distributing identical items. The formula for distributing \( n \) identical items into \( r \) distinct groups is given by:

\[ \binom{n + r - 1}{r - 1} \]

Here, we have \( n = 3 \) (the number of books) and \( r = 3 \) (the number of students).

For the physics books:
\[ \text{Ways to distribute physics books} = \binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10 \]

For the mathematics books:
\[ \text{Ways to distribute mathematics books} = \binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10 \]

### Step 2: Calculate the total ways to distribute both types of books.
Since the distributions of physics and mathematics books are independent, we multiply the two results:

\[ \text{Total ways} = 10 \times 10 = 100 \]

### Step 3: Calculate the cases where at least one student does not receive a book.
To find the number of distributions where at least one student does not receive a book, we can use complementary counting. We will first calculate the number of ways in which one specific student does not receive any books.

Assuming student A does not receive any books, we need to distribute the 3 physics books and 3 mathematics books between students B and C.

For the physics books:
\[ \text{Ways to distribute physics books to B and C} = \binom{3 + 2 - 1}{2 - 1} = \binom{4}{1} = 4 \]

For the mathematics books:
\[ \text{Ways to distribute mathematics books to B and C} = \binom{3 + 2 - 1}{2 - 1} = \binom{4}{1} = 4 \]

Thus, if student A does not receive any books, the total ways for students B and C is:
\[ \text{Ways for A not receiving} = 4 \times 4 = 16 \]

Since there are 3 students, the total cases where at least one student does not receive any books is:
\[ \text{Total cases where one student does not receive} = 3 \times 16 = 48 \]

### Step 4: Calculate the number of ways where each student gets at least one book.
To find the number of distributions where each student gets at least one book, we subtract the cases where at least one student does not receive any books from the total number of distributions:

\[ \text{Ways where each student gets at least one book} = \text{Total ways} - \text{Cases where one student does not receive} \]
\[ = 100 - 48 = 52 \]

### Final Answer
The number of ways of distributing 3 identical physics books and 3 identical mathematics books among three students such that each student gets at least one book is **52**.