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Neo1
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#1
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#1
Hello, this question I cant seem to get it....please help

The letters of the word POSSESSES are written on 9 cards, one on each card. The cards are shuffled and four of them are selected and arranged in a straight line.

(a) How many possibly selections are there of 4 letters.
(b) How many arrangements are there of 4 letters.

Please help me on these!
Thanks a lot
From John
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littlemissalex
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the question doesnt make sense.. if the cards each have a letter on them, then surely the possibility that the 4 cards have a letter on them is 1?
Maybe you misread the question or something cause i dont make sense of it.. maybe its just me lol
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Abdiel
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Do you know the answer? cos if its 63 or 1512 for part a) i think i can tell u how to do it. I could tell u anyway, but i might be wrong

Ab
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*LEGEND*
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This is actually a really hard question.
The answers are 12 and 115 from the s1 textbook. Any ideas anyone?
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Chewwy
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(Original post by alexandra)
the question doesnt make sense.. if the cards each have a letter on them, then surely the possibility that the 4 cards have a letter on them is 1?
Maybe you misread the question or something cause i dont make sense of it.. maybe its just me lol
it's just you.
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Chewwy
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may as well do this long hand. possibilities:

all Ss
3Ss and something else (P, O or E)
2Ss, 1E, 1 other (P,O or E)
2Ss, 1P, 1O
1S, 1E, 1 other (P,O,E)
EEOP

12 in total. now for each possibility work out the number of ways it can be ordered, and add them up.
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Abdiel
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okay, worked out the first part.

Basically, if it was just abcdefghi or any other combination of 9 different letters, then the total number of different combinations would be 4!*5!. If it was 6 cards dealt the no. of different combinations would be 6!*3!, if all the cards delt there would be a massive 9! combinations (362880).

Because there are 5 s, 2 e, 1 0 and 1 p, we divide the top nuber by 5!*2!*1!*1! (5!*2! cos 1! = 1) this is the number of possible ways that the available letters r gonna b arranged - and its gonna reduce the total cos there's less ways with less different letters. (4!*5!)/(5!*2!)= 12

pwhew. i'll have my dinner and then try and do the other one if no one else has by then.

NB: if u get this in the exam.. just turn the page to the next question

Ab
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xlinax
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I know it doesn't work out but i just wondered why you couldn't just do 9C4 seeing as 4 were chosen from 9?
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Chewwy
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(Original post by kornpie)
I know it doesn't work out but i just wondered why you couldn't just do 9C4 seeing as 4 were chosen from 9?
because some of the letters are repeated. for example, there are 5 different combinations you can choose which will all give you 4Ss.
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