- #1

mathmari

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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. ÌˆFinally, 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 ÌˆTens 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: