There are several tea cups in the kitchen, some with handle and the others without handles. The number of ways of selecting two cups without a handle and three with a handle is exactly 1200. What is the maximum possible number of cups in the kitchen?

Answer

To solve the problem, we need to determine the maximum number of tea cups in the kitchen, given the conditions regarding the selection of cups with and without handles.

### Step-by-Step Solution:

1. **Define Variables**:
Let \( x \) be the number of cups without handles and \( y \) be the number of cups with handles.

2. **Set Up the Equation**:
The number of ways to select 2 cups without handles is given by \( \binom{x}{2} \) and the number of ways to select 3 cups with handles is given by \( \binom{y}{3} \). According to the problem, we have:
\[ \binom{x}{2} \cdot \binom{y}{3} = 1200 \]

3. **Express the Combinations**:
The combinations can be expressed as:
\[ \binom{x}{2} = \frac{x(x-1)}{2} \quad \text{and} \quad \binom{y}{3} = \frac{y(y-1)(y-2)}{6} \]
Substituting these into the equation gives:
\[ \frac{x(x-1)}{2} \cdot \frac{y(y-1)(y-2)}{6} = 1200 \]
Simplifying this, we get:
\[ \frac{x(x-1) \cdot y(y-1)(y-2)}{12} = 1200 \]
Multiplying both sides by 12 results in:
\[ x(x-1) \cdot y(y-1)(y-2) = 14400 \]

4. **Factorization**:
We need to find integer values for \( x \) and \( y \) such that the product \( x(x-1) \cdot y(y-1)(y-2) = 14400 \).

5. **Finding Possible Values**:
We can factor 14400 to find suitable pairs of \( x(x-1) \) and \( y(y-1)(y-2) \). The prime factorization of 14400 is:
\[ 14400 = 2^6 \cdot 3^2 \cdot 5^2 \]
We will check different values for \( y \) and calculate \( y(y-1)(y-2) \) to find corresponding \( x(x-1) \).

6. **Testing Values for \( y \)**:
- For \( y = 4 \):
\[ y(y-1)(y-2) = 4 \cdot 3 \cdot 2 = 24 \]
Then,
\[ x(x-1) = \frac{14400}{24} = 600 \]
Solving \( x(x-1) = 600 \) gives \( x = 25 \) (since \( 25 \cdot 24 = 600 \)).
- For \( y = 5 \):
\[ y(y-1)(y-2) = 5 \cdot 4 \cdot 3 = 60 \]
Then,
\[ x(x-1) = \frac{14400}{60} = 240 \]
Solving \( x(x-1) = 240 \) gives \( x = 16 \) (since \( 16 \cdot 15 = 240 \)).
- For \( y = 6 \):
\[ y(y-1)(y-2) = 6 \cdot 5 \cdot 4 = 120 \]
Then,
\[ x(x-1) = \frac{14400}{120} = 120 \]
Solving \( x(x-1) = 120 \) gives \( x = 11 \) (since \( 11 \cdot 10 = 110 \)).
- For \( y = 10 \):
\[ y(y-1)(y-2) = 10 \cdot 9 \cdot 8 = 720 \]
Then,
\[ x(x-1) = \frac{14400}{720} = 20 \]
Solving \( x(x-1) = 20 \) gives \( x = 6 \) (since \( 6 \cdot 5 = 30 \)).

7. **Finding Maximum**:
The combinations we found are:
- \( (x, y) = (25, 4) \) → Total = 29
- \( (x, y) = (16, 5) \) → Total = 21
- \( (x, y) = (11, 6) \) → Total = 17
- \( (x, y) = (6, 10) \) → Total = 16

The maximum total is when \( x = 25 \) and \( y = 4 \), which gives:
\[ \text{Maximum number of cups} = 25 + 4 = 29 \]

### Final Answer:
The maximum possible number of cups in the kitchen is **29**.