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Math Help - Arranging the letters in Mississippi. Have the answer, can someone explain it?

  1. #1
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    Arranging the letters in Mississippi. Have the answer, can someone explain it?

    The eleven letters of the word MISSISSIPPI are scrambled and then arranged in some order.
    (1) What is the probability that the four I's are consecutive letters in the resulting arrangement?
    (2) What is the conditional probability that the four I's are consecutive (event A), given B , where B is the event that arrangement starts with M and ends with S?
    (3) What is conditional probability of A, as deŻned above, given C, where C is the event that the arrangements ends with four consecutive S's?

    I have the answers and have been sitting here trying to figure out the reasoning behind it, but I have had no luck. Can anybody help me out?

    The answers are as follows

    1.)
    8*[(7!)/(1!*4!*2!)]/[(11!)/1!*4!*4!*2!)] = 4/165

    2.)
    6*[C(5,2)]/[(9!)/(4!*3!*2!)] = 1/21

    3.)
    4*[C(3,2)]/[(7!)/(1!*2!*4!)] = 4/35
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  2. #2
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    Using subscripts MI_1S_1S_2I_2S_3S_4I_3P_1P_2I_4 now we have 11 different letters
    and there are 11! ways to rearrange those.

    In any of those rearrangements, how many ways can we leave the I’s fixed but rearrange their subscripts?
    Well it is 4!. Rright?
    So if we drop the subscripts on the I's we would have to divide by 4!.

    The same idea for any repeated letter. So the number of ways to rearrange the letters MISSISSIPPI is
    \frac{11!}{4!\cdot 4! \cdot 2!}

    Do you see this much?
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  3. #3
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    Arrangements with like items

    Hello babbagandu
    Quote Originally Posted by babbagandu View Post
    The eleven letters of the word MISSISSIPPI are scrambled and then arranged in some order.
    (1) What is the probability that the four I's are consecutive letters in the resulting arrangement?
    (2) What is the conditional probability that the four I's are consecutive (event A), given B , where B is the event that arrangement starts with M and ends with S?
    (3) What is conditional probability of A, as deŻned above, given C, where C is the event that the arrangements ends with four consecutive S's?

    I have the answers and have been sitting here trying to figure out the reasoning behind it, but I have had no luck. Can anybody help me out?

    The answers are as follows

    1.)
    8*[(7!)/(1!*4!*2!)]/[(11!)/1!*4!*4!*2!)] = 4/165

    2.)
    6*[C(5,2)]/[(9!)/(4!*3!*2!)] = 1/21

    3.)
    4*[C(3,2)]/[(7!)/(1!*2!*4!)] = 4/35
    Generalising Plato's result gives the number of arrangements of n items, where there are a_1 alike of the first kind, a_2 alike of the second kind, ... and so on, which is:
    \frac{n!}{a_1!a_2!...}
    1) To get the probability where all four I's are together, treat the block of 4 I's as a single item. So we have 8 items altogether, with 4 alike of the first kind (S) and 2 alike of the second kind (P). Using the general result above, then, the number of arrangements is
    \frac{8!}{4!2!}
    So the probability of this is
    \frac{8!}{4!2!}\div\frac{11!}{4!4!2!}=\frac{3}{165  }
    2) If the arrangement starts with M and ends with S, the remaining 9 letters have 4 alike of the first kind (I), 3 alike of the second kind (S) and 2 alike of the third kind (P). Using the general result above, then, there are therefore \frac{9!}{4!3!2!} arrangements of these letters.

    If we treat the 4 I's as a single block (as in question (1)), we have \frac{6!}{3!2!} arrangements. So the probability is
    \frac{6!}{3!2!}\div\frac{9!}{4!3!2!}=\frac{1}{21}
    You may wonder why the answer is given in an alternative form:
    6\times{^5C_2}\div\frac{9!}{4!3!2!}
    The reason is that, with 6 available 'slots' (1 slot for the block of 4 I's and 5 slots for the remaining 5 letters), there are 6 ways of selecting the position of the block of 4 I's, and then {^5C_2} ways of choosing the two slots to be occupied by the 2 P's; the 3 S's filling the remaining 3 slots.

    3) Use either of these two techniques to answer this question, where (once the 4 S's have been placed at the end) there are 7 items left with 4 alike of the first kind (I) and 2 alike of the second kind (P).

    Can you complete this one now?

    Grandad
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  4. #4
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    Quote Originally Posted by babbagandu View Post
    The eleven letters of the word MISSISSIPPI are scrambled and then arranged in some order.
    (1) What is the probability that the four I's are consecutive letters in the resulting arrangement?
    (2) What is the conditional probability that the four I's are consecutive (event A), given B , where B is the event that arrangement starts with M and ends with S?
    (3) What is conditional probability of A, as deŻned above, given C, where C is the event that the arrangements ends with four consecutive S's?

    I have the answers and have been sitting here trying to figure out the reasoning behind it, but I have had no luck. Can anybody help me out?

    The answers are as follows

    1.)
    8*[(7!)/(1!*4!*2!)]/[(11!)/1!*4!*4!*2!)] = 4/165

    2.)
    6*[C(5,2)]/[(9!)/(4!*3!*2!)] = 1/21

    3.)
    4*[C(3,2)]/[(7!)/(1!*2!*4!)] = 4/35
    Friend,

    I had the same question as yours which I Plato and Grandad helped me last Friday. In my case, I was given TENNESSEE. I spent many hours studying it and had tested it with computer to show my professor this morning. He agreed with my computer print-out. You might want to use the computer to help you study in this--it would help tremediously.

    Since I am still fresh with it. I thought, I ought to help you a little. I hope Plato and Grandad would not mind.

    1)
    (11!)/1!*4!*4!*2!) = total possible arrangements with repeating P's, S's, and I's
    8*[(7!)/(1!*4!*2!)] is the total arrangements assuming that there are 4 consecutive I’s plus 2s’s and one other letter, which the 2’s and the other letter arrange among themselves in 8 different ways. This answer which your book has provided is valid only if there are 2 other letters repeating, apart from the 4I's. This, I have run it through my professor before I typed it here.

    2.)
    6*[C(5,2)]/[(9!)/(4!*3!*2!)]

    Since M and W are assigned, you have only the remaining are 9 letter to arrange where you have 4I’s+3s’+2P’s to choose from, which = (9!)/(4!*3!*2!<--@)

    In you 7-letter, after 2 spaces are taken by M and S, you have 5 spaces left, and in the five spaces, you have 2 ways of placing your 4 consecutive I’s = [C(5,2)]
    After 4 spaces have been filled, you will have 6 ways of arranging 2 letters—2 ways for SS, 2 ways for PP, and 2 ways for SP. The repeats of SS and PP are taken care off by the 2! in the paragraph above, marked @.


    3.)
    4*[C(3,2)]/[(7!)/(1!*2!*4!)] = 4/35

    I cannot help on this. The question 3 is not clear.

    We are fellow students on this learning. Good luck!
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  5. #5
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    Thank you everyone for the detailed answers above. You all were a great help. Thanks!
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