Six people are playing Secret Santa for Christmas. They will each give one gift to someone, and each receive one gift from someone. They are not allowed to receive their own gift. How many different ways are there to exchange gifts?

To find the number of different ways to exchange gifts, we can use the concept of derangements.

A derangement is a permutation of a set where no element appears in its original position. In this case, we want to find derangements of the set of six people (A, B, C, D, E, F) such that no person receives their own gift.

To calculate the number of derangements, we can use the principle of inclusion-exclusion. We start with all possible permutations of the set (which is 6!, or 720).

Next, we need to subtract the number of permutations where at least one person receives their own gift. Let's calculate this:

- Number of permutations where A receives their own gift: There are 5 remaining people to give gifts to (B, C, D, E, F), so we have 5! possible permutations for them. For the remaining 4 people (B, C, D, E) to not receive their own gift, we can calculate the number of derangements for them, which is given by the formula D(4) = 4!(1 - 1/1! + 1/2! - 1/3! + 1/4!) = 9. By multiplying these two values together, we get 5! * 9 = 2,160.

- Similarly, there are 5! * 9 permutations where B, C, D, E, or F receives their own gift.

However, we have overcounted some scenarios. For example, we subtracted the permutations where both A and B receive their own gifts twice (once when considering A, and once when considering B). To correct this, we need to add back the permutations where two people receive their own gifts:

- Number of permutations where both A and B receive their own gifts: There are 4! ways to arrange the remaining 4 people (C, D, E, F) without any restrictions. For the remaining 3 people (C, D, E) to not receive their own gifts, we calculate D(3) = 3!(1 - 1/1! + 1/2!) = 2. By multiplying these two values together, we get 4! * 2 = 48.

- By symmetry, there are also 48 permutations where both C and D, C and E, C and F, D and E, D and F, or E and F receive their own gifts.

Finally, we need to subtract the permutations where three people receive their own gifts (A, B, and C; A, B, and D; etc.), which can be calculated in a similar manner.

- Number of permutations where A, B, and C receive their own gifts: There are 3! ways to arrange the remaining 3 people (D, E, F). For the remaining person (D) to not receive their own gift, we calculate D(1) = 1!(1 - 1/1!) = 0. By multiplying these two values together, we get 3! * 0 = 0.

- By symmetry, there are also 0 permutations where A, B, and D; A, B, and E; A, B, and F; A, C, and D; A, C, and E; A, C, and F; B, C, and D; B, C, and E; B, C, and F; B, D, and E; B, D, and F; or C, D, and E receive their own gifts.

Now, we can apply the principle of inclusion-exclusion to find the final count:

Total derangements = 6! - (5! * 9) + (4! * 2) - (3! * 0)
= 720 - (5 * 2,160) + (24 * 2) - 0
= 720 - 10,800 + 48 - 0
= -9,032

The negative value indicates that there are no valid arrangements where no one receives their own gift. Therefore, there are no different ways to exchange gifts among the six people in this scenario, while respecting the given constraints.