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 when there are four types of biscuits available and the person must take at least one biscuit of each type, we can follow these steps:

### Step 1: Understand the Problem
We have 4 types of biscuits, and we need to buy a total of 10 biscuits, ensuring that we take at least one biscuit of each type.

### Step 2: Allocate One Biscuit to Each Type
Since we need to take at least one biscuit of each type, we first allocate 1 biscuit to each of the 4 types. This means we will distribute 4 biscuits (1 for each type) right away.

### Step 3: Calculate Remaining Biscuits
After allocating 1 biscuit to each of the 4 types, we have:
\[ 10 - 4 = 6 \text{ biscuits remaining} \]
Now, we need to distribute these 6 remaining biscuits among the 4 types.

### Step 4: Use the Stars and Bars Theorem
The problem now reduces to finding the number of ways to distribute 6 identical biscuits (remaining biscuits) into 4 distinct types (biscuit varieties).

According to the "Stars and Bars" theorem, the number of ways to distribute \( n \) identical objects into \( r \) distinct groups is given by the formula:
\[ \binom{n + r - 1}{r - 1} \]
In our case, \( n = 6 \) (remaining biscuits) and \( r = 4 \) (types of biscuits).

### Step 5: Apply the Formula
Substituting the values into the formula:
\[ \binom{6 + 4 - 1}{4 - 1} = \binom{9}{3} \]

### Step 6: Calculate the Binomial Coefficient
Now, we calculate \( \binom{9}{3} \):
\[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3! \cdot 6!} \]
Calculating this gives:
\[ = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84 \]

### Conclusion
Thus, the total number of ways a person can buy 10 biscuits, ensuring at least one of each type, is **84**.

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