Gift exchange activity with n people and k gift(probability)

Question

I have the following question:

There is a gift exchange activity among n persons. In this activity, everyone prepares k gifts which have the same packaging. All gifts are then put in a non-transparent box and mixed well so that they are not distinguishable. People take gifts in turn. In the turn of a person, they takes k gifts from the box uniformly at random one by one (without replacement).
(a) If n = 6 and k = 1, what is the probability that everyone gets a gift not prepared by themself? (Please keep 8 significant digits.)
(b) If n = 6 and k = 2, what is the probability that everyone gets the two gifts prepared by themself? (Please keep 8 significant digits.)
(c) If n = 6 and k = 2, what is the probability that everyone gets at least one gift not prepared by themself? (Please keep 8 significant digits)
(d) Assume k = 2. In the turn of person X, if the first gift X takes is from person Y, what is the probability that the other gift X takes is also from Y ?

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I start part a with the inclusion-exclusion principle using the formula of $$\sum_{i=0}^n \frac {(-1)^k}{k!}$$ For part a, there are 6 people and each will bring 1 gift.
So it will be the case that $$\sum_{i=0}^6 \frac {(-1)^k}{k!}$$ Then we can get $$1-\frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!}+... + \frac{1}{6!} =\frac{53}{144}$$ By keeping 8 significant digits, we get the answer $$0.36805556$$ For part b, I think it is quite similar to part a but how to handle the case that everyone brings 2 gifts? I try to consider all possible cases for selecting two gifts which is $$ 12C_2 \times10C_2\times 8C_2 \times6C_2 \times4C_2 \times 2C_2$$ After simplifying the formula, I get the formula: $$ \frac {1}{64} \times 12!$$ I try to plug in the formula and get $$\frac {1}{64}\sum_{i=0}^{12} \frac {(-1)^k}{k!}$$ and I got the probability of $0.00574116$, is this the probability of no one getting their own gift? If yes, I got the probability for part c which is $$ 1- 0.00574116 = 0.99425188$$ but I don't know if it is correct or not, could anyone please help?
and any approach to do part b and part d?

Answer 1

Part (c):

Let $A_1$ be the event that our first person receives both of their prepared gifts. Similarly define $A_2, A_3,\dots,A_6$ for the second, third, ... sixth persons respectively.

You are tasked with finding $\Pr(A_1^c\cap A_2^c\cap \cdots \cap A_6^c)$, that is... the probability that all six persons received at least one gift not prepared by themselves. To do this, view this indirectly as $1 - \Pr(A_1\cup A_2\cup \cdots \cup A_6)$

This can be expanded by inclusion-exclusion.

$1 - \Pr(A_1)-\Pr(A_2)-\dots+\Pr(A_1\cap A_2)+\dots-\Pr(A_1\cap A_2\cap A_3)\pm\dots$

Each term of which should be straightforward to find, the last term of which is precisely what you were tasked with finding for part (b).

$\Pr(A_1\cap A_2)$ for instance, the probability the first person got both of their prepared gifts and same for the second person. There are $\binom{12}{2}\binom{10}{2}$ ways to pick the gifts for the first and second person (where we don't care about order in which they receive the gifts, and we care only about tracking the gifts given to these two people.) Exactly one of those ways has it with each person getting their own gifts only, so $\Pr(A_1\cap A_2) = \frac{1}{\binom{12}{2}\binom{10}{2}}$

Continue in this fashion and complete the problem.

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As for part (d)... without loss of generality just imagine that person X is the first person to start taking gifts from the gift pile. After having already gotten a gift from person Y, there remain $2n-1$ gifts in the pile, exactly one of which is the other gift from Y.