Game Night : Number of combinations

[mathmari]

Hey!

Sarah, Alex and Mary meet at Tom for a game night.

a) The first game is a board game in which six game colors (red, green, yellow, blue, violet, orange) are possible. Each of the players chooses a color in turn. (Each color can only be assigned once.) How many different color distributions are there for the four players?

b) During a break in the game, Tom goes to the kitchen and gets a drink for everyone. They can choose between a box of beer, water, soda, cola, apple juice and orange juice.
How many combination possibilities are there for the four reactors that he brings from the kitchen? (Anyone who subsequently acquires the respective attraction should be left unconsidered.)

c) Next, double-head is played. The 48 (distinguishable) double-header cards are shuffled well and each player receives 12 cards. How many possibilities are there for the 12 cards that Mary receives? How many possibilities are there in total, to distribute the cards?

d) Finally, a kind of lottery is played in order to know who has to get up. Thereby everyone writes down a five-digit number consisting only of the digits 1 to 4. ̈Finally, four pieces of paper are mixed with the digits 1 to 4 and the digits of the number are drawn one after the other (with the notes being returned). First the one and finally the ̈Tens of thousands. If you have the fewest agreements, you have to clean the room. How many possibilities for the five-digit number are available?
In addition, refer to the urn model used, a justification for the choice, the number $n$ of the balls in the urn and the number $k$ of the drawn.

a) Is it like the problem where we chooses a ball from a box without replacement?

So the number of color distributions is equal to $\binom{6}{4}=15$, right?

b) Is it like the problem where we chooses a ball from a box with replacement?

So the number of possibilities is equal to $6^4$, right?

c) Is it like the problem where we chooses a ball from a box without replacement?

So the number of color distributions is equal to $\binom{48}{12}$, right?

d) Is it like the problem where we chooses a ball from a box with replacement?

So the number of possibilities is equal to $4^5$, right? :unsure:

 

[I like Serena]

a) Is it like the problem where we chooses a ball from a box without replacement?

So the number of color distributions is equal to $\binom{6}{4}=15$, right?

Hey mathmari!

It is like that problem yes, but we still need to consider if we get an ordered or an unordered set from the box.

b) Is it like the problem where we chooses a ball from a box with replacement?

So the number of possibilities is equal to $6^4$, right?

Yes. (Nod)

c) Is it like the problem where we chooses a ball from a box without replacement?

So the number of color distributions is equal to $\binom{48}{12}$, right?

Not exactly.
The ordering matters again.
It matters who receives a specific card, but it does not matter in which order a single person receives their cards.

d) Is it like the problem where we chooses a ball from a box with replacement?

So the number of possibilities is equal to $4^5$, right?

Yep. (Nod)

 

[mathmari]

It is like that problem yes, but we still need to consider if we get an ordered or an unordered set from the box.

In this case we don't get an ordered set, do we? :unsure:

Not exactly.
The ordering matters again.
It matters who receives a specific card, but it does not matter in which order a single person receives their cards.

Why does the ordering matter? I got stuck right now. :unsure:

 

[I like Serena]

In this case we don't get an ordered set, do we?

An ordered set is for instance (red, green, yellow, blue) while the corresponding unordered set is {red, green, yellow, blue}.
Suppose Sarah chooses red and Alex chooses green, is that different from the situation that Sarah chooses green and Alex chooses red or not? (Wondering)

Why does the ordering matter? I got stuck right now. :unsure:

Consider card number 1.
If Sarah gets it, we have a different outcome than if Alex gets it.
However, suppose the first two cards that Sarah gets are first card 1 and then card 2.
Is that different or the same as when she gets first card 2 and then card 1?

There are 48! ordered ways to distribute the cards.
But it is irrelevant in which order Sarah receives her cards exactly.
She can receive her cards in 12! different ways that are equivalent. So we need to divide by 12! to compensate for the fact that the ordering of Sarah's cards does not matter.

 

[I like Serena]

Actually, (c) asks 2 questions:

c) Next, double-head is played. The 48 (distinguishable) double-header cards are shuffled well and each player receives 12 cards. How many possibilities are there for the 12 cards that Mary receives? How many possibilities are there in total, to distribute the cards?

It appears you actually answered the first question and not the second.
There are indeed $\binom{48}{12}$ possibilities for the cards that Mary receives. (Nod)

That leaves the question how many possibilities there are in total, to distribute the cards. (Sweating)

 

[mathmari]

An ordered set is for instance (red, green, yellow, blue) while the corresponding unordered set is {red, green, yellow, blue}.
Suppose Sarah chooses red and Alex chooses green, is that different from the situation that Sarah chooses green and Alex chooses red or not? (Wondering)

As I understand the the statement (a) the ordering matters, doesn't it?

Does this mean then that the number of ways is equal to $\binom{6}{4}\cdot 4!=360$ or we use the formula $\frac{n!}{(n-k)!}=\frac{6!}{(6-4)!}=\frac{6!}{2!}=360$, right?

:unsure:

 

[mathmari]

That leaves the question how many possibilities there are in total, to distribute the cards. (Sweating)

Is this equal to $\frac{48!}{12!}$ ? :unsure:

 

[I like Serena]

As I understand the the statement (a) the ordering matters, doesn't it?

Does this mean then that the number of ways is equal to $\binom{6}{4}\cdot 4!=360$ or we use the formula $\frac{n!}{(n-k)!}=\frac{6!}{(6-4)!}=\frac{6!}{2!}=360$, right?

Yep. (Sun)

Is this equal to $\frac{48!}{12!}$ ?

That is the number of ways where the ordering of Sarah's cards does not matter, but where the ordering of Alex's, Mary's, and Tom's cards does matter when they shouldn't. (Worried)

We can do this in two ways. We can first consider the cards that Mary receives, and then consider which of the remaining cards Alex receives, and so on.
Or we can calculate the total number of ordered distributions, and then divide by 12! for each person for who the order does not matter.

 

[mathmari]

That is the number of ways where the ordering of Sarah's cards does not matter, but where the ordering of Alex's, Mary's, and Tom's cards does matter when they shouldn't. (Worried)

We can do this in two ways. We can first consider the cards that Mary receives, and then consider which of the remaining cards Alex receives, and so on.
Or we can calculate the total number of ordered distributions, and then divide by 12! for each person for who the order does not matter.

By the second way you mean $$\frac{48!}{12!\cdot 12!\cdot 12!\cdot 12!}$$ ? :unsure:

 

[I like Serena]

By the second way you mean $$\frac{48!}{12!\cdot 12!\cdot 12!\cdot 12!}$$ ?

Yes. (Nod)

And the first alternative is $\binom{48}{12}\cdot \binom{36}{12}\cdot \binom{24}{12}\cdot \binom{12}{12}$, which is the same thing. (Nerd)

 

[mathmari]

At (b) and (d) do we have "choosing a ball from a box with replacement and ordering matters" ? :unsure:

 

[mathmari]

Yes. (Nod)

And the first alternative is $\binom{48}{12}\cdot \binom{36}{12}\cdot \binom{24}{12}\cdot \binom{12}{12}$, which is the same thing. (Nerd)

At (c) do we have two different kind of problems?

First when we consider the possibilities for the 12 cards that Mary receives is equivalent to choosing a ball from box without replacement and without that the ordering matters.

Then when we consider the possibilities in total, to distribute the cards is equivalent to choosing a ball from box without replacement and the ordering matters?

:unsure:

 

[I like Serena]

At (b) and (d) do we have "choosing a ball from a box with replacement and ordering matters" ?

Yes. (Nod)

At (c) do we have two different kind of problems?

First when we consider the possibilities for the 12 cards that Mary receives is equivalent to choosing a ball from box without replacement and without that the ordering matters.

Then when we consider the possibilities in total, to distribute the cards is equivalent to choosing a ball from box without replacement and the ordering matters?

Correct. (Nod)

 

[mathmari]

Yes. (Nod)

At (b) the ordering matters because it is important if for example Sarah takes the water and Mary the orange juice or the other way around, right?
At (d) again the ordering matters because we have to know which of the four has the most correctly number since this person has to clean the room, tight?

:unsure:

Correct. (Nod)

Could you explain to me why it is like that? It is not very clear to me. :unsure:

 

[I like Serena]

At (b) the ordering matters because it is important if for example Sarah takes the water and Mary the orange juice or the other way around, right?

Yep. (Nod)

At (d) again the ordering matters because we have to know which of the four has the most correctly number since this person has to clean the room, tight?

The ordering matters because the numbers are compared in order: "First the one and finally the ̈Tens of thousands."

Could you explain to me why it is like that? It is not very clear to me.

It seemed you already said how it was.
What is not clear? (Wondering)

Anyway, first we draw 12 cards for Mary and give them to her.
We might shuffle those 12 cards and it won't matter. So they are unordered.

Next we draw 12 cards for Alex, and we cannot swap those cards with Mary or it's a different possibility.
Now the order does matter.

So we first draw 12 cards from the 48 for Mary.
And then we draw 12 cards from the remaining 36 for Alex.
And so on.