The number of ways of distributing six identical mathematics books and six identical physics books among three students such that each student gets atleast one mathematics book and atleast one physics book is `(5.5!)/k`, then k is..

Answer

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

### Step 1: Distributing Mathematics Books
We need to ensure that each of the three students (let's call them A, B, and C) receives at least one mathematics book. To achieve this, we can initially give one mathematics book to each student. This means we have distributed 3 mathematics books, leaving us with 3 mathematics books to distribute freely among the three students.

### Step 2: Formulate the Distribution Equation
After giving one mathematics book to each student, we need to distribute the remaining 3 mathematics books. We can represent this distribution as:
\[ A + B + C = 3 \]
where \( A \), \( B \), and \( C \) represent the number of additional mathematics books received by students A, B, and C respectively.

### Step 3: Use the Stars and Bars Theorem
The problem of distributing \( n \) identical items (in this case, the remaining mathematics books) into \( r \) distinct groups (the students) can be solved using the stars and bars theorem. The formula for this is:
\[ \binom{n + r - 1}{r - 1} \]
In our case, \( n = 3 \) (the remaining mathematics books) and \( r = 3 \) (the students). Thus, we have:
\[ \binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} \]

### Step 4: Distributing Physics Books
We follow the same procedure for the physics books. Each student must receive at least one physics book, so we again give one physics book to each student, leaving us with 3 physics books to distribute freely.

Using the same equation:
\[ A + B + C = 3 \]
we again apply the stars and bars theorem:
\[ \binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} \]

### Step 5: Calculate Total Ways
Since the distributions of mathematics and physics books are independent, we multiply the number of ways to distribute the mathematics books by the number of ways to distribute the physics books:
\[ \text{Total Ways} = \binom{5}{2} \times \binom{5}{2} \]

### Step 6: Calculate the Binomial Coefficient
Calculating \( \binom{5}{2} \):
\[ \binom{5}{2} = \frac{5!}{2! \cdot 3!} = \frac{5 \times 4}{2 \times 1} = 10 \]

Thus, the total number of ways becomes:
\[ \text{Total Ways} = 10 \times 10 = 100 \]

### Step 7: Relate to Given Expression
We are given that the total number of ways can also be expressed as:
\[ \frac{5 \times 5!}{k} \]
Setting the two expressions equal gives:
\[ 100 = \frac{5 \times 5!}{k} \]

### Step 8: Solve for \( k \)
To find \( k \), we first calculate \( 5! \):
\[ 5! = 120 \]
Thus:
\[ 100 = \frac{5 \times 120}{k} \]
\[ 100k = 600 \]
\[ k = \frac{600}{100} = 6 \]

### Final Answer
The value of \( k \) is:
\[ \boxed{6} \]