Problem 137

Let set A represent the 2 students who play chess but not Go, set B represent the 3 students who play Go but not chess, and set C represent the 4 students who can play both. The method to choose 2 students for the competitions can be broken down into the following 4 cases:
- Select 1 person from set A for the chess competition and 1 person from set B for the Go competition.
- Select 1 person from set C for the chess competition and 1 person from set B for the Go competition.
- Select 1 person from set C for the Go competition and 1 person from set A for the chess competition.
- Select 2 people from set C, one for each competition.

Now, let's calculate each scenario:

First case (A for chess, B for Go):
The number of ways to select 1 person from A and 1 from B is:
$$|A| \times |B| = 2 \times 3 = 6$$

Second case (C for chess, B for Go):
The number of ways to select 1 person from C and 1 from B is:
$$|C| \times |B| = 4 \times 3 = 12$$

Third case (A for chess, C for Go):
The number of ways to select 1 person from A and 1 from C is:
$$|A| \times |C| = 2 \times 4 = 8$$

Fourth case (2 people from C):
The number of ways to select 2 people from C for the two competitions is equivalent to the number of ways to choose 2 different people from a group of 4, which can be calculated as a combination:
$$\binom{|C|}{2} = \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$$

Adding all these up, the total number of distinct ways to choose the 2 students is:
$$6 + 12 + 8 + 6 = 32$$

Therefore, there are $\boxed{32}$ different ways to choose the students for the chess and Go competitions.