To determine the number of social relationships in a group of four people, we can think of each relationship as a connection between two people. This is a classic problem in combinatorics, where we need to find the number of ways to choose 2 people out of 4 to form a relationship.

The formula for combinations is given by:

$$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$

where $ n $ is the total number of people, and $ k $ is the number of people in each group (in this case, 2).

For our problem:
- $ n = 4 $ (the total number of people)
- $ k = 2 $ (since a relationship involves 2 people)

Plugging these values into the formula, we get:

$$ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2! \cdot 2!} $$

Calculating the factorials, we have:

$$ 4! = 4 \times 3 \times 2 \times 1 = 24 $$
$$ 2! = 2 \times 1 = 2 $$

So,

$$ \binom{4}{2} = \frac{24}{2 \times 2} = \frac{24}{4} = 6 $$

Therefore, the number of social relationships linking the members of the group is 6.

The correct answer is:
C. six

Question

To determine the number of social relationships in a group of four people, we can think of each relationship as a connection between two people. This is a classic problem in combinatorics, where we need to find the number of ways to choose 2 people out of 4 to form a relationship.

The formula for combinations is given by:

$$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$

where $ n $ is the total number of people, and $ k $ is the number of people in each group (in this case, 2).

For our problem:
- $ n = 4 $ (the total number of people)
- $ k = 2 $ (since a relationship involves 2 people)

Plugging these values into the formula, we get:

$$ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2! \cdot 2!} $$

Calculating the factorials, we have:

$$ 4! = 4 \times 3 \times 2 \times 1 = 24 $$
$$ 2! = 2 \times 1 = 2 $$

So,

$$ \binom{4}{2} = \frac{24}{2 \times 2} = \frac{24}{4} = 6 $$

Therefore, the number of social relationships linking the members of the group is 6.

The correct answer is:
C. six

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