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    Hey everyone. How many of you have gave paper 6 ( stats ) today? I gave paper 62 and it wasn't really that hard except a permutations question . What about you guys? How was the paper for you? I'm very happy because since P 12 was so easy so I expected this to be harder :P
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    same here, definitely got that permutation and combinations question wrong but other than that everything else was pretty easy. grade boundaries must be really high.
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    Yeah true the paper was indeed too good to be true . I think i got the permutation question right although im not quite sure.
    I'm just waiting for the solutions to leak online
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    the permutations question was easy!

    i got one little query though-what was your answer to no.2 where they ask for the probability for time taken between 5-10 minutes? it was of only one mark.
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    (Original post by SamihaTahsin)
    the permutations question was easy!

    i got one little query though-what was your answer to no.2 where they ask for the probability for time taken between 5-10 minutes? it was of only one mark.
    We just had to subtract the 2 probabilities. So the ans was 0.72 . What was your ans to the question of "exactly 6 letters between 2 W" mine was 20160 :/
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    (Original post by Some one)
    We just had to subtract the 2 probabilities. So the ans was 0.72 . What was your ans to the question of "exactly 6 letters between 2 W" mine was 20160 :/
    I got 3720 but I am not really sure about it since I considered them asking us an 8 letter alphabet. But my friend said it should be a 10 letter one and if that is the case, it would become different.
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    (Original post by Coloneltreavy)
    I got 3720 but I am not really sure about it since I considered them asking us an 8 letter alphabet. But my friend said it should be a 10 letter one and if that is the case, it would become different.
    Yea we had to consider a 10 letter alphabet not 8 .
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    (Original post by Some one)
    We just had to subtract the 2 probabilities. So the ans was 0.72 . What was your ans to the question of "exactly 6 letters between 2 W" mine was 20160 :/

    don't remember the exact value but it was pretty huge like the one you quoted. and i am quite sure mine is right, finished all the others real quick and had 25 mins to do the permutations.
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    (Original post by Some one)
    Yea we had to consider a 10 letter alphabet not 8 .
    Then 22320
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    (Original post by Coloneltreavy)
    Then 22320
    Nah I don't think so.
    We had to first consider 3Ls between 2W then 2L between 2W then 1L then add them up.
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    (Original post by Some one)
    Nah I don't think so.
    We had to first consider 3Ls between 2W then 2L between 2W then 1L then add them up.
    This is exactly what I did.

    With three Ls, you need to add a combination of 3 alphabets so 5C3
    Multiply it by 6P6 because 6 numbers and divide by 3!
    Repeat wit 2 L: 5C4*6P6/2!
    With 1L: 5C5*6P6
    We can consider W to W as a whole and 2 letters are left out so multiply the sum of previous example with 3! (This, I missed)
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    (Original post by Coloneltreavy)
    This is exactly what I did.

    With three Ls, you need to add a combination of 3 alphabets so 5C3
    Multiply it by 6P6 because 6 numbers and divide by 3!
    Repeat wit 2 L: 5C4*6P6/2!
    With 1L: 5C5*6P6
    We can consider W to W as a whole and 2 letters are left out so multiply the sum of previous example with 3! (This, I missed)
    Yes you're right except that for 1L we have to multiply by 3!/2! Beacuse there are 2L outside now.
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    This is how I did it:
    So first 3Ls between the the 2 W so we need three more letters from 5 more so 5C3, then arranging so
    5C3*(6!/3!)*3! = 7200
    Then 2Ls : 5C4*(6!/2!)*3! = 10800
    Then 1L : 5C5*6!*(3!/2!) = 2160
    Total arrangements = 7200+10800+2160=20160.
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    (Original post by Some one)
    This is how I did it:
    So first 3Ls between the the 2 W so we need three more letters from 5 more so 5C3, then arranging so
    5C3*(6!/3!)*3! = 7200
    Then 2Ls : 5C4*(6!/2!)*3! = 10800
    Then 1L : 5C5*6!*(3!/2!) = 2160
    Total arrangements = 7200+10800+2160=20160.
    You're right. This makes sense. I wonder if they will give me any marks :')
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    I bet you will score 2/4 marks for method (which isn't too bad)
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    (Original post by Some one)
    I bet you will score 2/4 marks for method (which isn't too bad)
    I dont know man. CIE is pretty strict on marking. I would assume them to give 1 mark for the 3L thing, another for 2L, and one more for L. Last mark for correct answer. That's how they usually do.
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    Hey everyone, follow these links
    https://justpastpapers.com/9709_s16_qp_62/
    https://justpastpapers.com/9709_s16_ms_62/
 
 
 
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