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    why for this combinations question is 3! not used because the order obviously matters because it is a code

    the answer is 4 * 3 *2

    this is for question 2 i
    http://mei.org.uk/files/papers/s108ja_leqa.pdf

    why was 3! used for 4C3 but not for 4 * 3*2
    i don't really understand combinations, when can they be used?
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    (Original post by esmeralda123)
    why for this combinations question is 3! not used because the order obviously matters because it is a code

    the answer is 4 * 3 *2

    this is for question 2 i
    http://mei.org.uk/files/papers/s108ja_leqa.pdf

    why was 3! used for 4C3 but not for 4 * 3*2
    i don't really understand combinations, when can they be used?
    Combinations do suck.

    There are 4 ways of choosing the first letter, 3 the second and 2 the third (without replacement, as the letters have to be distinct) hence its 4*3*2.

    That's not particularly convincing, so it may be worth revisiting the textbook or finding something online.
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    (Original post by esmeralda123)
    why for this combinations question is 3! not used because the order obviously matters because it is a code

    the answer is 4 * 3 *2

    this is for question 2 i
    http://mei.org.uk/files/papers/s108ja_leqa.pdf

    why was 3! used for 4C3 but not for 4 * 3*2
    i don't really understand combinations, when can they be used?
    For part (i) there can be no repetitions. So for the first letter you have 4 options, then for second letter you have 3 options, and for last letter you have 2 options. Thus by the product rule for counting, there are 4 \cdot 3 \cdot 2 = 24 possibilities. Alternatively, this is given by the function 4\mathrm{P}3 = \frac{4!}{(4-3)!}

    It cannot be 3! because this number simply represents the number of ways you can arrange 3 letters. It doesn't say anything about the fact that you can choose from 4 objects.
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    (Original post by RDKGames)
    Doesn't say that the order matters, so it doesn't.

    For part (i) there can be no repetitions. So for the first letter you have 4 options, then for second letter you have 3 options, and for last letter you have 2 options. Thus by the product rule for counting, there are 4 \cdot 3 \cdot 2 = 24 possibilities. Alternatively, this is given by the function 4\mathrm{P}3 = \frac{4!}{(4-3)!}

    It cannot be 3! because this number simply represents the number of ways you can arrange 3 letters. It doesn't say anything about the fact that you can choose from 4 objects.
    but aren't permutations used when order matters?
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    they should have made the question clearer by including

    "for instance the codes AGC and CAG are distinct"

    :mad:
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    (Original post by esmeralda123)
    but aren't permutations used when order matters?
    Yes. Got distracted and didnt mean to say that it doesnt matter. So it does, hence you can use permutation function.

    But yeah it would’ve been nice if they made this clearer
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    (Original post by RDKGames)
    Yes. Got distracted and didnt mean to say that it doesnt matter. So it does, hence you can use permutation function.

    But yeah it would’ve been nice if they made this clearer
    so why couldn't 3! be used if order did matter?
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    (Original post by esmeralda123)
    so why couldn't 3! be used if order did matter?
    It doesnt say anything about the fact that you have 4 objects to choose from.

    You can say that it is 4C3*3! but this is precisely just 4P3
 
 
 
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