Exploring Possibilities with 6 Pairs of Gloves

[evinda]

Hey again! (Wasntme)

We have $6$ pairs of gants,all of them have a different colour.
We share the $6$ right gants randomly to $6$ people,then we share the $6$ left gants randomly to the same people.
Find the possibility:

• Nobody has a pair of the same colour
• Everybody has a pair of the same colour
• Only one gets a pair of the same colour
• at least $2$ get a pair of the same colour

That's what I have tried (Blush):Let $D_n=|\{ \sigma \in S_n: \sigma(i) \neq i \forall i \in [n]\}|$

• $6! \cdot D_6$
• $6!$
• $6! \cdot 5!$
• $6!-D_2$

Could you tell me if it is right?

 

[I like Serena]

Hey again! (Wasntme)

We have $6$ pairs of gants,all of them have a different colour.
We share the $6$ right gants randomly to $6$ people,then we share the $6$ left gants randomly to the same people.
Find the possibility:

• Nobody has a pair of the same colour
• Everybody has a pair of the same colour
• Only one gets a pair of the same colour
• at least $2$ get a pair of the same colour

Hey! (Mmm)

Hmm. What do you mean by "possibility"?
Do you mean the number of possibilities?
Or do you mean the probability? (Wondering)

And what is a "gant"? (Thinking)

That's what I have tried (Blush):Let $D_n=|\{ \sigma \in S_n: \sigma(i) \neq i \forall i \in [n]\}|$

Could you tell me if it is right?

• $6! \cdot D_6$
• $6!$

Right and right! (Sun)

• $6! \cdot 5!$

Hmm... suppose the first person gets a pair.
How many possibilities to divide the remaining gants? (Thinking)

• $6!-D_2$

Hmm... $D_2$ is the number of possibilities that 2 gants are divided differently... which is 1 possibility.
I don't think that will work! (Doh)

 

[soroban]

Hello, evinda!

We have 6 pairs of gants, each pair a different color.
We share the 6 right gants randomly to 6 people,
then we share the 6 left gants randomly to the same people.

Find the probability that:
(a) Nobody has a pair of the same color.
(b) Everyone has pair of the same color.
(c) Exactly one gets a pair of the same color.
(d) At least two get a pair of the same color.

A derangement is an distribution
in which no one has a matching pair.

Let $$d(n)$$ = number of derangements of $$n$$ objects.

There are: $$6! = 720$$ possible outcomes.

(a) $$P(\text{No matches}) \:=\:\frac{d(6)}{6!} \:=\:\frac{265}{720} \:=\:\frac{53}{144}$$(b) There is one way in which there are 6 matches.
$$P(\text{6 matches}) \:=\:\frac{1}{720}$$(c) There are 6 choices for the person with the match.
The other 5 gants must be deranged.
$$P(\text{1 match}) \:=\;\frac{6\cdot d(5)}{6!} \:=\:\frac{6\cdot 44}{720} \:=\:\frac{11}{30}$$(d) The opposite of "at least two" is "0 or 1".
We already know: $$P(0) =\tfrac{265}{720},\;P(1) = \tfrac{264}{720}$$
Hence: $$P(\text{0 or 1}) \:=\:\tfrac{265}{720}+\tfrac{264}{720} \:=\:\tfrac{529}{720}$$
Therefore: .$$P(\text{2 or more}) \:=\:1 - \frac{529}{720} \:=\:\frac{191}{720}$$

 

[Nono713]

Also I think "gant" means "glove" here. Probably translation error :p

 

[blue_raver22]

Hello Wasntme,

Your approach to solving this problem is a good start. However, I would suggest using a different notation to represent the possibilities. Instead of using $D_n$ to represent the number of derangements (permutations with no fixed points), you can use $N_n$ to represent the number of permutations where nobody has a pair of the same colour. Then, the possibilities can be represented as follows:

1. $N_6$: This represents the number of permutations where nobody has a pair of the same colour. This can be calculated using the principle of inclusion-exclusion, where we subtract the number of permutations with at least one person having a pair of the same colour from the total number of permutations ($6!$). This can be written as $6! - \binom{6}{1} \cdot N_5 + \binom{6}{2} \cdot N_4 - \binom{6}{3} \cdot N_3 + \binom{6}{4} \cdot N_2 - \binom{6}{5} \cdot N_1 + \binom{6}{6} \cdot N_0$. Here, $N_5$ represents the number of permutations where at least one person has a pair of the same colour, but not everyone has a pair of the same colour. Similarly, $N_4$ represents the number of permutations where at least two people have a pair of the same colour, but not everyone has a pair of the same colour, and so on. $N_0$ represents the number of permutations where everyone has a pair of the same colour.

2. $N_0$: This represents the number of permutations where everyone has a pair of the same colour. This can be calculated as $\binom{6}{1} \cdot 1 \cdot N_5 - \binom{6}{2} \cdot 2 \cdot N_4 + \binom{6}{3} \cdot 3 \cdot N_3 - \binom{6}{4} \cdot 4 \cdot N_2 + \binom{6}{5} \cdot 5 \cdot N_1 - \binom{6}{6} \cdot 6 \cdot N_0$. Here, $N_5$ represents the number of ways to choose one colour for all six pairs of gloves,