in Combinatory recategorized by
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4 votes
4 votes
Consider $5$ cards, each has a distinct value from the set $\{2,3,4,5,6\},$ so there are $5$ different values, and we put them face down on the table. There are $5$ players and each player is given a number from $2$ to $6,$ such that no two players have the same number. Each player is given a card by a dealer. A player loses if the value of the given card coincides with the value that player has. If no player loses, then the dealer loses.
How many ways are there so that the dealer loses?
in Combinatory recategorized by
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1 Answer

2 votes
2 votes

This is just a derangement problem of $5$ elements. $\text{D5} = 44.$

Video Solution:

https://youtu.be/tqjuxfutFHg?t=3124

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2 Comments

Why are we not considering the number of ways that 5 number can be given to that 5 players i.e. 5! ?

Isn’t that need to be included, cause for each 5 distinct arrangement, there will be D5 = 44 ways the dealer can loose.
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The question is “In how many ways the Dealer Loses”.

Dealer loses only when No player loses.

No player loses means No player gets a card of value equal to their own value. Hence, we are looking for Derangement here.

See Video Solution:

https://youtu.be/tqjuxfutFHg?t=3124

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