Guest amikan Posted May 12, 2005 Share Posted May 12, 2005 A deck of n cards is laid out in a row on the table. Cards of a second deck with n cards are placed one by one at random on top of the first set of cards. You get m points if there are m matches between the first and second decks. a) How many ways are there to get 2 points if n = 4? b) What is the probability of getting 7 points if n = 9? I got stuck in this question...could anybody help me out??Thanks!! Link to comment Share on other sites More sharing options...
dan19_83 Posted May 12, 2005 Share Posted May 12, 2005 when you say n=4, does this mean you pulled 4 random cards from a normal 52 card deck? Link to comment Share on other sites More sharing options...
kriminal99 Posted May 24, 2005 Share Posted May 24, 2005 A deck of n cards is laid out in a row on the table. Cards of a second deck with n cards are placed one by one at random on top of the first set of cards.You get m points if there are m matches between the first and second decks. a) How many ways are there to get 2 points if n = 4? b) What is the probability of getting 7 points if n = 9? I got stuck in this question...could anybody help me out??Thanks!! Thats a hard one. To begin with, it seems like a little clarification is needed to determine what is "different". Is a match of the same cards at different locations different? Also is it ok to assume that the first deck of n cards contains the same cards in it as the second deck of n cards? Link to comment Share on other sites More sharing options...
uncool Posted May 24, 2005 Share Posted May 24, 2005 Each of them mean that there are two cards that are different. Let us renumber and rearrange the cards so that the first n-2 cards are in order, and the last two cards are switched in one deck. We then must figure out how many ways the cards can be renumbered and rearranged. There are n!/2 ways of renumbering the cards (n! because there are n cards, /2 because switching the order of the two switched cards is not allowed). There are n(n-1) ways of placing the cards (n-1 for the first switched card to be placed, n for the second). So the total number of ways is n*(n-1)*n!/2. For the first, it is 4*3*24/2 = 144 For the second, it is 9*8*362880/2 possibilities. Divide by n!^2. 9*8/2*9! = 1/2*7! = 1/10080 Someone check if I'm right? -Uncool- Link to comment Share on other sites More sharing options...
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