Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    16
    ReputationRep:
    Hi all,

    I don't understand why 3(d) requires the "*6" at the end of the calculation?:

    Calculation:
    0.16/0.8 * 0.56/0.8 * 0.08/0.8 *6

    Answer:
    0.084

    Thank you for your help!
    Attached Images
     
    • Very Important Poster
    Offline

    21
    ReputationRep:
    Very Important Poster
    (Original post by londoncricket)
    Hi all,

    I don't understand why 3(d) requires the "*6" at the end of the calculation?:

    Calculation:
    0.16/0.8 * 0.56/0.8 * 0.08/0.8 *6

    Answer:
    0.084

    Thank you for your help!
    There are \frac{3!}{1! \times1! \times 1!} ways of arranging those days, which is the same as 3!, which is 6.

    It's just like having 3 balls RBG in a bag, each of a different colour, and there are 6 ways of taking them out (RBG, RGB, GBR, GRB, BRG, BGR).
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by SeanFM)
    There are \frac{3!}{1! \times1! \times 1!} ways of arranging those days, which is the same as 3!, which is 6.

    It's just like having 3 balls RBG in a bag, each of a different colour, and there are 6 ways of taking them out (RBG, RGB, GBR, GRB, BRG, BGR).

    Thanks for your prompt reply!

    So would it be x!, where x corresponds to the number of possible combinations?

    Thank you!
    • Very Important Poster
    Offline

    21
    ReputationRep:
    Very Important Poster
    (Original post by londoncricket)
    Thanks for your prompt reply!

    So would it be x!, where x corresponds to the number of possible combinations?

    Thank you!
    Not quite

    If you have n objects, r of one type and n-r of the other, then the number of ways that you can arrange them is \frac{n!}{(n-r)!r!)}.

    More generally, if you have n objects, a of one type, b of another, c.... such that a + b + c + .... = n, then the number of ways you can arrange them is \frac{n!}{a!b!c!...} and in this case, it's just a,b and c with a = b = c = 1.
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by SeanFM)
    Not quite

    If you have n objects, r of one type and n-r of the other, then the number of ways that you can arrange them is \frac{n!}{(n-r)!r!)}.

    More generally, if you have n objects, a of one type, b of another, c.... such that a + b + c + .... = n, then the number of ways you can arrange them is \frac{n!}{a!b!c!...} and in this case, it's just a,b and c with a = b = c = 1.
    Thank you for that.

    I am not sure as to whether I fully understand your response. In this case, why does a = b = c = 1? Also, why would I do this in this question: where are the combinations here?

    Also, do I simply stick (n!)/(a!)/(b!)/(c!) on to the end of the 0.16/0.8 * 0.56/0.8 * 0.08/0.8? Or should this be done in replacement of the calculation from above?

    Thank you!
    • Very Important Poster
    Offline

    21
    ReputationRep:
    Very Important Poster
    (Original post by londoncricket)
    Thank you for that.

    I am not sure as to whether I fully understand your response. In this case, why does a = b = c = 1? Also, why would I do this in this question: where are the combinations here?

    Also, do I simply stick (n!)/(a!)/(b!)/(c!) on to the end of the 0.16/0.8 * 0.56/0.8 * 0.08/0.8? Or should this be done in replacement of the calculation from above?

    Thank you!
    The three different types of objects, a, b and c in this case, is that it arrives early (a), arrives on time (b) or arrives late (c) and from the formula, there are 6 different ways of ordering them.

    To see this, let's look at some orders. The first one could be early, second on time, third late. The probability of this is
    0.16/0.8 * 0.56/0.8 * 0.08/0.8 as you say. (The order in which you multiply them may be different but it does not matter).

    Another order could be the reverse - first day is late, second is on time, third is early. The probability is the same as the first one. So combine this one and the first one and that gives you 2 * 0.16/0.8 * 0.56/0.8 * 0.08/0.8

    And you can get four more orders (from the order or just trying RBG as in a previous post), and their probabilities will be equal so it gives you 6 * ..., and that 6 can come directly from the formula.

    You can stick the formula on at the end, or calculate it elsewhere and put in 6. (If you have never seen the formula before, ask your teacher about it).
    Offline

    22
    ReputationRep:
    Have you heard of the binomial distribution?
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by SeanFM)
    The three different types of objects, a, b and c in this case, is that it arrives early (a), arrives on time (b) or arrives late (c) and from the formula, there are 6 different ways of ordering them.

    To see this, let's look at some orders. The first one could be early, second on time, third late. The probability of this is
    0.16/0.8 * 0.56/0.8 * 0.08/0.8 as you say. (The order in which you multiply them may be different but it does not matter).

    Another order could be the reverse - first day is late, second is on time, third is early. The probability is the same as the first one. So combine this one and the first one and that gives you 2 * 0.16/0.8 * 0.56/0.8 * 0.08/0.8

    And you can get four more orders (from the order or just trying RBG as in a previous post), and their probabilities will be equal so it gives you 6 * ..., and that 6 can come directly from the formula.

    You can stick the formula on at the end, or calculate it elsewhere and put in 6. (If you have never seen the formula before, ask your teacher about it).
    Ahh I got it! Thank you so much!

    In the attachment, why would you not use the factorial formula on (b). I understand that you would use it on (c).


    (Original post by Zacken)
    Have you heard of the binomial distribution?
    Yes I have. I understand that this and the factorial formula have something to do with Pascal's triangle and the nCr function on the calculator.
    Attached Images
     
    • Very Important Poster
    Offline

    21
    ReputationRep:
    Very Important Poster
    (Original post by londoncricket)
    Ahh I got it! Thank you so much!

    In the attachment, why would you not use the factorial formula on (b). I understand that you would use it on (c).




    Yes I have. I understand that this and the factorial formula have something to do with Pascal's triangle and the nCr function on the calculator.
    Think of it like tossing a coin. There is only one way of getting HH, one way of TT but two ways of HT. (In this case, think of both labour as both heads).

    (you're also only looking at one object, so it would be 2!/2! = 1 way.
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by SeanFM)
    Think of it like tossing a coin. There is only one way of getting HH, one way of TT but two ways of HT. (In this case, think of both labour as both heads).

    (you're also only looking at one object, so it would be 2!/2! = 1 way.
    I think I understand. You multiply it by the number of ways that the combination can be made. For example in the P(Labour, Labour) one, it is still being multiplied by 1!, which is just 1.

    Am I right in thinking this?

    Thank you.
    • Very Important Poster
    Offline

    21
    ReputationRep:
    Very Important Poster
    (Original post by londoncricket)
    I think I understand. You multiply it by the number of ways that the combination can be made. For example in the P(Labour, Labour) one, it is still being multiplied by 1!, which is just 1.

    Am I right in thinking this?

    Thank you.
    Correct it is really 2!/2!*0! Where 0! = 1.
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by SeanFM)
    Correct it is really 2!/2!*0! Where 0! = 1.
    Great! So would I do this for every question in which there are a possible number of combinations that is greater than one?
    Offline

    0
    ReputationRep:
    (Original post by londoncricket)
    Hi all,

    I don't understand why 3(d) requires the "*6" at the end of the calculation?:

    Calculation:
    0.16/0.8 * 0.56/0.8 * 0.08/0.8 *6

    Answer:
    0.084

    Thank you for your help!
    This 6 is needed because the 3 events (A-B-C) could happen in any combination: ABC or ACB or BAC or BCA or CAB or CBA. I hope this is helpful.
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by MARIOUSKA)
    This 6 is needed because the 3 events (A-B-C) could happen in any combination: ABC or ACB or BAC or BCA or CAB or CBA. I hope this is helpful.
    Thank you for that!
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by SeanFM)
    Correct it is really 2!/2!*0! Where 0! = 1.
    Please view this attachment.

    5(b)(ii)

    My method and answer:
    0.85*0.60*0.45 + 0.85*0.55*0.40 + 0.60*0.55*0.15 * 2!
    = 0.516

    Correct method and answer:
    0.85*0.60*0.45 + 0.85*0.55*0.40 + 0.60*0.55*0.15
    = 0.466

    Why would I not put the "* 2!" here? Is there not a combination here? Rhys could purchase those two items in two different orders?

    Thank you for your help!
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.