The first problem asks: A pair of distinct dice are thrown. Find the number of possible outcomes if the sum of the results is equal to 5 or 6. The second problem asks: There are 3 types of taxis for traveling to/from a village near a town. When traveling to town, Pu Sok can take any taxi or walk. When returning home, he decides to take a taxi. How many ways can he travel to and from? The third problem asks: There are 6 men and 5 women playing badminton in pairs. How many ways can a man and a woman form a pair?

1. Problem Description

The first problem asks: A pair of distinct dice are thrown. Find the number of possible outcomes if the sum of the results is equal to 5 or

6. The second problem asks: There are 3 types of taxis for traveling to/from a village near a town. When traveling to town, Pu Sok can take any taxi or walk. When returning home, he decides to take a taxi. How many ways can he travel to and from?

The third problem asks: There are 6 men and 5 women playing badminton in pairs. How many ways can a man and a woman form a pair?

2. Solution Steps

Problem 1:

Let $(x, y)$ be the outcome of the two dice, where $x$ and $y$ are the results of the first and second dice, respectively. We want to find the number of pairs such that $x + y = 5$ or $x + y = 6$.

For $x + y = 5$, the possible pairs are $(1, 4), (2, 3), (3, 2), (4, 1)$. There are 4 such pairs.

For $x + y = 6$, the possible pairs are $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$. There are 5 such pairs.

However, the problem states that the two dice are distinct. So, we can have distinct outcomes in both cases. In the case of sum 5 the two dice are distinct and are ok. In the case of sum 6 the dice are also distinct and therefore are ok.

Since the sum can be 5 or 6, we add the number of pairs for each case: $4 + 5 = 9$.

Problem 2:

When going to town, Pu Sok has 3 taxi choices or walking, so there are $3 + 1 = 4$ options.

When returning home, Pu Sok only takes a taxi, so there are 3 taxi choices.

The total number of ways to travel to and from is the product of the number of options for each trip: $4 \times 3 = 12$.

Problem 3:

There are 6 men and 5 women. We want to find the number of ways to form a man-woman pair.

For each of the 6 men, there are 5 possible women to pair with. Therefore, the number of ways to form a pair is $6 \times 5 = 30$.

3. Final Answer

1. 9

2. 12

3. 30