In a bakery four types of biscuits are available. In how many ways a person can buy 10 biscuits if he decide to take atleast one biscuit of each variety?

Answer

To solve the problem of how many ways a person can buy 10 biscuits from 4 types, ensuring that they take at least one biscuit of each variety, we can follow these steps:

### Step 1: Define the variables
Let the number of biscuits of each type be represented as follows:
- Let \( a \) be the number of type 1 biscuits.
- Let \( b \) be the number of type 2 biscuits.
- Let \( c \) be the number of type 3 biscuits.
- Let \( d \) be the number of type 4 biscuits.

### Step 2: Set up the equation
Since the person must buy a total of 10 biscuits, we can express this as:
\[ a + b + c + d = 10 \]

### Step 3: Account for the condition of at least one biscuit of each type
Since the person must take at least one biscuit of each type, we can redefine the variables to account for this condition:
- Let \( a' = a - 1 \)
- Let \( b' = b - 1 \)
- Let \( c' = c - 1 \)
- Let \( d' = d - 1 \)

Now, \( a', b', c', d' \) can be zero or more (non-negative integers). Thus, we can rewrite the equation as:
\[ (a' + 1) + (b' + 1) + (c' + 1) + (d' + 1) = 10 \]
This simplifies to:
\[ a' + b' + c' + d' = 6 \]

### Step 4: Use the stars and bars theorem
Now, we need to find the number of non-negative integer solutions to the equation:
\[ a' + b' + c' + d' = 6 \]

According to the stars and bars theorem, the number of solutions to the equation \( x_1 + x_2 + ... + x_k = n \) in non-negative integers is given by:
\[ \binom{n + k - 1}{k - 1} \]

In our case, \( n = 6 \) and \( k = 4 \) (since we have 4 types of biscuits). Thus, we need to calculate:
\[ \binom{6 + 4 - 1}{4 - 1} = \binom{9}{3} \]

### Step 5: Calculate \( \binom{9}{3} \)
Now, we compute \( \binom{9}{3} \):
\[ \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84 \]

### Final Answer
Thus, the total number of ways a person can buy 10 biscuits, ensuring at least one of each type, is:
\[ \boxed{84} \]