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Math Help - combinations

  1. #1
    Gul
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    combinations

    1)How many four letter words can be formed from the word proportion?

    2)How many committees of 9 people can be chosen from 12, if two certain people must never be in the same committee?

    1) p is repeated twice, as is r, and o is repeated three times

    so would it be 10C4 / (2!2!3!) to get rid of the double ups?

    2) 1,2,3,4,5,6,7,8,9,10,11,12

    lets say 1 is taken out, u have 11C9, and the same if two was taken out 11C9, and for the case where they are both together:10C9??

    I need some help with this one.
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  2. #2
    Senior Member pankaj's Avatar
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    Quote Originally Posted by Gul View Post
    1)How many four letter words can be formed from the word proportion?

    2)How many committees of 9 people can be chosen from 12, if two certain people must never be in the same committee?
    1)2 P's,3 O's,2 R's,1 T,1 I,1 N

    Case 1 All letters are different= \binom{6}{4}(4!)

    Case 2 Two letters same and two different= \binom{3}{1}.\binom{5}{2}.\frac{4!}{2!}

    Case 3 Two letters same and two letters same= \binom{3}{2}.\frac{4!}{2!2!}

    Case 4 Three letters same and one different== \binom{5}{1}.\frac{4!}{3!}

    Total number of words=Add up all the cases

    2)No. of ways to make the committee= \binom{12}{9}

    No. of ways in which the committee can be formed so that the two persons are in the same committee= \binom{10}{7}

    No. of ways in which the committee can be made so that the two persons are not in the same committee= \binom{12}{9}-\binom{10}{7}
    Last edited by pankaj; June 9th 2009 at 07:26 AM.
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  3. #3
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    Hello, Gul!

    2) How many committees of 9 people can be chosen from 12 people,
    if two certain people must never be in the same committee?
    There is a back-door approach to this problem . . .


    With no restrictions, there are: . _{12}C_9 \:=\:220 possible committees.


    Let the two anti-social people be A and B.

    How many committees have both A and B together?
    Place A and B on the committee.
    Choose the other 7 members from the remaining 10 people: . _{10}C_7 \:=\:120 ways.
    . . Hence, there are 120 committees with both A and B.

    Therefore, there are: . 220 - 120 \:=\:\boxed{100} committees
    . . in which A and B are not together.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    This problem can be solved head-on . . . It just takes longer.


    There are three cases to consider.

    (1) A is on the committee, hence B is not.
    . . .We choose the other 8 members from the remaining 10 people.
    . . .There are: . _{10}C_8 \:=\:45 ways.

    (2) B is on the committee, hence A is not.
    . . .We choose the other 8 members from the remaining 10 people.
    . . .There are: . _{10}C_8 \:=\:45 ways.

    (3) Neither A nor B is on the committee.
    . . .Choose the 9 members from the other 10 people.
    . . .There are: . _{10}C_9 \:=\:10 ways.

    Therefore, there are: . 45 + 45 + 10 \:=\:\boxed{100} committees
    . . in which A and B are not together.

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  4. #4
    Gul
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    Thanks guys. Very helpful
    But I don't understand why you have to multiply by 4!, 4!/2! etc?

    I thought about them again, for the first one:

    I have 6 letters available pick one: 6C1*1C1, then 5C1*1C1, then 4C1*1C1, then 3C1*1C1, so thats 6*5*4*3 = 360
    that's the same as 6C4*4! = 360

    For the second situation:two same, two different

    (3C1*2C2*5C1*1C1*4C1*1C1)/2! ?

    ^^3 ways of picking pairs, pick both cards from that pair, 5 types of letters available pick one, then four pick one?
    Last edited by Gul; June 10th 2009 at 12:25 AM. Reason: thinking about it
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  5. #5
    Senior Member pankaj's Avatar
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    my dear,you are also required to arrange the letters to form the words.Simply selection will not do
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