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    Hi,
    If anyone is awake could you explain question 8? i dont get any of it..


    http://www.ocr.org.uk/Images/242458-...atistics-1.pdf

    Why the hell is i) 5C2 over 8C4?? I get the 8C4 but not why its 5C2..
    Also ii) and iii) please
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    Hello.

    8/c/i is 5C2 over 8C4 because you're assuming that Kathy and David have been picked already.

    Put it this way: you have already taken Harpreet out of the equation, so now there are only 7 people left. Out of these 7, you have already established that Kathy and David must be in your group, so we can ignore them. Now there are only 5 people left. Out of these 5 people, you have to choose 2, so 5C2.

    8/c/ii/ is 5!*2!. This question can be graphically represented as so: H xxxxx KD

    From here, we can see that H stays the same and has 1! many positions. All the people in between H and KD can be arranged 5! ways. KD can be arranged 2!. Finally, KD can be at any position within x such that both H x KD xxxx is possible, for instance. There are six ways this can happen (write these graphically if you need to). Thus, our final answer is (1!)*5!*2!*6.

    8/c/iii/ is slightly more complicated to explain, but, again, this can be represented graphically. Let's first see how many arrangements of D, H, and K are possible:

    DHK // DKH // HDK // HKD // KDH // KHD

    If you have trouble with this, just order them by the first alphabetical letter first (in this case, D, then H, then K, and repeat). Now, the question says that H cannot stand between K and D (otherwise, he would not be next to them, he would be between them). Thus, we can take out KHD and DHK, leaving just 4 possibilities. Now, this set can be considered one block (let's call it A), like so:

    xxxxxA, xxxxAx, xxxAxx...

    It is easy to see, then, that there are six ways to arrange the five other people, as well as our block. This leaves 4*6!, which is our final answer.
 
 
 
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