To test a tea expert's claim, 10 cups of tea are prepared: in 5 cups milk is added before tea leaves and in the other 5 cups tea leaves are added before milk. In how many different ways can these 10 cups be presented in a line to the expert?

Introduction / Context:
This question is about counting sequences with a fixed number of two types of items. There are 10 cups, 5 of type A (milk first) and 5 of type B (tea first), and we are only interested in the order in which these types appear, not in any physical differences between cups of the same type. This leads directly to a combinations with repetition style count, often called a multinomial situation with two groups.

Given Data / Assumptions:

• Total cups = 10.
• 5 cups are of type A (milk added first).
• 5 cups are of type B (tea leaves added first).
• Cups of the same type are indistinguishable from each other for the purpose of ordering.
• We want the number of distinct sequences of these 10 cups based on their type pattern.

Concept / Approach:
We are essentially forming a 10 position sequence made up of 5 identical A items and 5 identical B items. The number of distinct sequences of n objects where you have n1 of one type and n2 of another type is given by n! / (n1! * n2!). In this situation, n = 10, n1 = 5 and n2 = 5.

Step-by-Step Solution:
Total positions to fill = 10.Number of A type cups = 5, number of B type cups = 5.Number of distinct sequences = 10! / (5! * 5!).Compute numerator: 10! = 3628800.Compute denominator: 5! * 5! = 120 * 120 = 14400.Divide: 3628800 / 14400 = 252.So there are 252 distinct ways to present the 10 cups.

Verification / Alternative check:
You could also think of selecting which 5 positions among the 10 are occupied by the A type cups. That selection can be done in 10C5 ways.10C5 = 252, which matches our earlier calculation because 10C5 = 10! / (5! * 5!).

Why Other Options Are Wrong:
210 and 290 are near 252 but come from incorrect combination calculations or arithmetic errors.340 is significantly larger than the correct value and does not correspond to 10Ck for any k.

Common Pitfalls:
Treating each cup as distinct and using 10! directly, which overcounts many identical type arrangements.Forgetting that only the pattern of types A and B matters, not which specific physical cup occurs at each position.Confusing 10C5 with 5C10, although numerically they are equal, the interpretation must be clear.

Final Answer:
The 10 cups can be presented in 252 different ways.