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  1. #1
    asc
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    ways

    How many ways are there to arrange 4 americans, 3 russians and 5 chinese into a queue, in such a way that no nationality forms a single consecutive block?
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  2. #2
    Member SengNee's Avatar
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    (12C4)(8C3)(5C5)-(9C1+10C1+8C1)
    =166320-27
    =166293

    I not sure....
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  3. #3
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    Quote Originally Posted by asc View Post
    How many ways are there to arrange 4 americans, 3 russians and 5 chinese into a queue, in such a way that no nationality forms a single consecutive block?
    Complementary Counting

    I will assume the people are not identical

    Find the number of ways in which they do form a consecutive block and then you can just subtract this number from the total.

    I think it will run into a few cases:
    Total number of ways in which americans form a block = 9!
    Total number of ways in which russians form a block = 10!
    Total number of ways in which chinese form a block = 8!

    But we have counted some solutions multiple number of times.

    Total number of ways in which russians and americans form a block = 7!
    Total number of ways in which russians and chinese form a block = 6!
    Total number of ways in which americans and chinese form a block = 5!
    Total number of ways in which russians, americans and chinese form a block = 3!

    So total number of ways: 9!+10!+8! - (7!+6!+5!) + 3! (You can also say ,"by principle of inclusion and exclusion")
    Now subtract it from 12!
    So
    Answer:12! - (9!+10!+8! - (7!+6!+5!) + 3!)
    I got 474975474 ways

    P.S:I am most likely wrong. I find counting questions hard always. But I have tried working this problem. Tell me if you did not understand something, most likely the step is wrong
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  4. #4
    Super Member wingless's Avatar
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    Quote Originally Posted by Isomorphism View Post
    Total number of ways in which americans form a block = 9!
    Total number of ways in which russians form a block = 10!
    Total number of ways in which chinese form a block = 8!
    They must be 9! 4!, 10! 3! and 8! 5!. You took americans as a group but probably forgot that there are 4! american groups. Same goes for all.
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  5. #5
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    I do not think that either of the two replies is correct.
    Let A be the event that all the Americans are together as a block. Then use R and C stand for the same for the other two groups. Now we want the number \left| {A^c R^c C^c } \right|, that is not A, not R and not C (none of the groups forms a single consecutive block).

    You will want to justify each of the following.
    The number of ways that all the American forms a single consecutive block is \left| A \right| = \left( {4!} \right)\left( {9!} \right), WHY?.
    The number of ways that all the American and all of the Russians are together as a consecutive block is \left| AR \right| = \left( {4!} \right) \left( {3!} \right)\left( {7!} \right), WHY?.

    The number of ways that at least one of the national groups forms a single consecutive block is:
    \left| {A \cup R \cup C} \right| = \left| A \right| + \left| B \right| + \left| C \right| - \left| {AR} \right| - \left| {AC} \right| - \left| {RC} \right| + \left| {ARC} \right|.

    Now you want \left| {A^c R^c C^c } \right| = 12! - \left| {A \cup R \cup C} \right|.
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