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    Hi guys, basically, there's a permutation question that i dont understand, the q is:

    Suppose repetitions are not permitted.
    (a) Find the number of three-digit numbers that can be formed from the six digits 2,3,4,5,6,7 and 9.

    (b) how many of them are less than 400?

    (c) how many of them are even.

    ANSWER:
    (A) P(6,3) - CAN SOMEONE EXPLAIN THAT TO ME?

    (B) 2*5*4


    (C) 2*5*4


    I would really appreciate if someone could explain the answers to me


    thanks!!
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    (Original post by mabz123)
    Hi guys, basically, there's a permutation question that i dont understand, the q is:

    Suppose repetitions are not permitted.
    (a) Find the number of three-digit numbers that can be formed from the six digits 2,3,4,5,6,7 and 9.

    (b) how many of them are less than 400?

    (c) how many of them are even.

    ANSWER:
    (A) P(6,3) - CAN SOMEONE EXPLAIN THAT TO ME?

    (B) 2*5*4


    (C) 2*5*4


    I would really appreciate if someone could explain the answers to me


    thanks!!
    You have listed 7 digits not 6. Assuming there are 6 digits:

    Think of the choices you have for each digit. a) For the first digit there are 6 choices, the second digit then has 5 choices since this digit cannot be the same as the first. Then for the third digit there are 4 choices.

    Multiplying these together you have 6 x 5 x 4 = 120 choices for the 3 digit number. This is the same as P(6,3).

    Have a go at b) and c) now.

    EDIT : The answer to c) looks like a mistake. Check you've written the question correctly.
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    For c) it might be helpful to choose the digits in reverse order. But I get 3*5*4 bit confused. 3 being the number of even numbers from the list to choose from.
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    thank you both for replying to the thread, much appreciated.

    @notnek, thanks for the explanation, i get it now.
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    (Original post by notnek)
    ...
    i have another question, i was wondering if you would be able to help me with it.

    The question is:

    A women student is to answer 10 out of 13 questions. Find the number of her choices where she must answer:

    (a) the first two questions
    (b) the first or second question but not both
    (c) Exactly three out of the first 5 question
    (d) atleast 3 of the first 5 questions


    The answer is:
    (a) 165
    (b) 110
    (c) 80
    (d) 276

    i'd really appreciate if someone could explain these answers to me.

    thanks!
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    (Original post by mabz123)
    i have another question, i was wondering if you would be able to help me with it.

    The question is:

    A women student is to answer 10 out of 13 questions. Find the number of her choices where she must answer:

    (a) the first two questions
    (b) the first or second question but not both
    (c) Exactly three out of the first 5 question
    (d) atleast 3 of the first 5 questions


    The answer is:
    (a) 165
    (b) 110
    (c) 80
    (d) 276

    i'd really appreciate if someone could explain these answers to me.

    thanks!
    a) She has no choice for the first 2 questions. There are 8 more questions that she can choose out of 11 choices. Order mattered in your first question but doesn't here (choosing question 1 then 2 then 3 is no different from choosing 2 then 3 then 1). So this is a combination problem and the answer is 11 C 8 (11 choose 8) = 165.

    See if you can do the rest of the questions now.
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    (Original post by notnek)
    a) She has no choice for the first 2 questions. There are 8 more questions that she can choose out of 11 choices. Order mattered in your first question but doesn't here (choosing question 1 then 2 then 3 is no different from choosing 2 then 3 then 1). So this is a combination problem and the answer is 11 C 8 (11 choose 8) = 165.

    See if you can do the rest of the questions now.
    ok, so i got the answer to b, and i got it by doing 2*C(11,9)
    BUT, im still confused on how to do part c, please help!
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    (Original post by mabz123)
    ok, so i got the answer to b, and i got it by doing 2*C(11,9)
    BUT, im still confused on how to do part c, please help!
    There are C(5,3) for the first 3 questions. Since there must be exactly 3 chosen from the first 5, the remaining 7 questions can only be chosen from 13-5 = 8 choices.

    Can you finish it?
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    (Original post by mabz123)
    Hi guys, basically, there's a permutation question that i dont understand, the q is:

    Suppose repetitions are not permitted.
    (a) Find the number of three-digit numbers that can be formed from the six digits 2,3,4,5,6,7 and 9.

    (b) how many of them are less than 400?

    (c) how many of them are even.

    ANSWER:
    (A) P(6,3) - CAN SOMEONE EXPLAIN THAT TO ME?

    (B) 2*5*4


    (C) 2*5*4


    I would really appreciate if someone could explain the answers to me


    thanks!!
    The answer to (c) should be 3 x 5 x 6
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    (Original post by brianeverit)
    The answer to (c) should be 3 x 5 x 6
    I got 3x5x4. The third digit is chosen from the 3 even numbers, the second digit is then chosen from the remaining 5 (even or odd) numbers and the first digit chosen from the remaining (even or odd) 4 numbers.
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    (Original post by steve44)
    I got 3x5x4. The third digit is chosen from the 3 even numbers, the second digit is then chosen from the remaining 5 (even or odd) numbers and the first digit chosen from the remaining (even or odd) 4 numbers.
    My answer was because you had listed 7 integers 2,3,4,5,6,7 & 9
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    (Original post by brianeverit)
    My answer was because you had listed 7 integers 2,3,4,5,6,7 & 9
    Oh right. I'm not the OP by the way - it's not my list. And I think we concluded near the start of the thread (see notnek's first post) that there should only be 6 numbers there, because otherwise the answers to (a) (and (b)) would be wrong. Personally I just automatically omitted the 9 but didn't say so...now come to think of it if I'd omitted one of the even numbers from the list (maybe either the 4 or the 6 so as not to change the answer to part (b)) the answer stated in the question for part (c) wouldn't be wrong...
 
 
 
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