In a shop, there are five types of ice-creams available. A child buys six ice-creams.
Statement-1: The number of different ways the child can buy the six ice-creams is `.^(10)C_(4)`.
Statement-2: The number of different ways the child can buy six ice-creams is equal to the number of different ways to arranging 6A's and 4B's in a row.

Options

A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1
B. Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1
C. Statement-1 is true, statement-2 is false
D. Statement-1 is false, statement-2 is true

Explanation

To solve the problem of how many different ways a child can buy six ice-creams from five types available, we can use the concept of combinations with repetition.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the number of ways to distribute 6 identical ice-creams among 5 different types. This can be modeled using the "stars and bars" theorem in combinatorics.

2. **Setting Up the Equation**:
Let \( x_1, x_2, x_3, x_4, x_5 \) represent the number of ice-creams of each type bought by the child. We need to solve the equation:
\[ x_1 + x_2 + x_3 + x_4 + x_5 = 6 \]
where \( x_i \) (for \( i = 1, 2, 3, 4, 5 \)) are non-negative integers.

3. **Applying the Stars and Bars Theorem**:
According to the stars and bars theorem, the number of non-negative integral solutions to the equation \( x_1 + x_2 + ... + x_r = n \) is given by:
\[ \binom{n + r - 1}{r - 1} \]
Here, \( n = 6 \) (the total number of ice-creams) and \( r = 5 \) (the types of ice-creams).

4. **Calculating the Combinations**:
Plugging in the values:
\[ \binom{6 + 5 - 1}{5 - 1} = \binom{10}{4} \]

5. **Conclusion for Statement 1**:
Thus, the number of different ways the child can buy the six ice-creams is indeed \( \binom{10}{4} \), confirming Statement 1 is true.

6. **Understanding Statement 2**:
Now, we need to verify Statement 2, which states that the number of different ways to buy six ice-creams is equal to the number of ways to arrange 6 A's and 4 B's in a row.

7. **Arranging A's and B's**:
If we consider the 6 ice-creams as A's and the remaining 4 as B's, we are looking for the number of distinct arrangements of the sequence consisting of 6 A's and 4 B's.

8. **Using the Formula for Arrangements**:
The number of distinct arrangements of \( n \) items where there are \( p \) of one kind and \( q \) of another kind is given by:
\[ \frac{n!}{p! \cdot q!} \]
Here, \( n = 10 \) (total items), \( p = 6 \) (A's), and \( q = 4 \) (B's). Thus, we have:
\[ \frac{10!}{6! \cdot 4!} = \binom{10}{4} \]

9. **Conclusion for Statement 2**:
Therefore, the number of ways to arrange 6 A's and 4 B's is also \( \binom{10}{4} \), confirming Statement 2 is also true.

### Final Answer:
Both statements are true, and the number of different ways the child can buy the six ice-creams is \( \binom{10}{4} \).