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Math Help - Permutations help me please

  1. #1
    Junior Member Misa-Campo's Avatar
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    Permutations help me please

    Hi i need urgent help please , i don't know where this topic belongs on this forum since it seems kinda miscellaneous. Sorry to ask so much for a 1st post.

    Q1) A car registration consists of 3 letters following by a number between 000 to 999

    a) how many car numbers are possible?
    b) How many possible car number if the 3 letters are the same?

    Q2)
    If repetitions aren't allowed, how many even numbers less than 400 can be made using the digits 1,2,3,4,5,6,7?

    Q3) 5 travelers arrive in a town where there are 5 hotels

    a) how many different arrangements if there are no restrictions on where the travelers wish to stay?

    b) Suppose 2 travelers are husband and wife and must go to the same hotel. how many different accommodation arrangements are there if the other 3 can go to any of the other hotels?

    Please give me some tips for these questions, i dont have a clue on how to do these, thanks alot.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Misa-Campo View Post
    Hi i need urgent help please , i don't know where this topic belongs on this forum since it seems kinda miscellaneous. Sorry to ask so much for a 1st post.
    this is actually in probability/statistics forum.. anyway, you don't have to post it there since any topics can be posted here.

    Quote Originally Posted by Misa-Campo View Post
    Q1) A car registration consists of 3 letters following by a number between 000 to 999

    a) how many car numbers are possible?
    b) How many possible car number if the 3 letters are the same?
    _ _ _ - _ _ _
    a)
    how many letters can be placed in each of the first three blanks? there are 26 letters right? therefore, you can choose any of the 26 for the first blank, 26 for the second, and 26 for the third..
    for the next 3 blanks, how many choices do you have? 10 each (0,1,...,9)
    therefore, the number cars of possible is the product..
    26 x 26 x 16 x 10 x 10 x 10

    b) if a letter has been chosen to one of the first three blanks, (whether the first, second or third), you would have no choice to other 2 blanks of the first three.. so you would have
    26 x 1 x 1 x 10 x 10 x 10
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Misa-Campo View Post
    Q2)
    If repetitions aren't allowed, how many even numbers less than 400 can be made using the digits 1,2,3,4,5,6,7?
    how many choices do you have in the hundreds digit?
    how many in the tens digit?
    in ones digit?

    take note you need a number less that 400, and there are no repetitions..
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Misa-Campo View Post
    Q3) 5 travelers arrive in a town where there are 5 hotels

    a) how many different arrangements if there are no restrictions on where the travelers wish to stay?
    case1: all five stay together in one hotel. how many ways?
    case2: 4 stay together on 1 hotel, the other 1 is isolated. how many ways?
    case3: 3 stay together on 1 hotel, 1 stays alone and also the other 1 stays alone.
    case4: 3 stay together on 1 hotel, 2 stays together on other hotel.
    case5: 2 stay together on 1 hotel, 3 distributes themselves to other hotels.
    case6: 2 stay together on 1 hotel, another 2 stay together on another hotel and 1 isolates.
    case4*: 2 stay together on 1 hotel, the other 3 stay on another hotel. (but this is the same as the case4. so this is not included anymore.)
    case7: all 5 distributes themselves without accompanying others. (each 1 stays alone on one hotel)
    case2*: 1 stays alone, 4 stay together. (same as case2.)
    case3*: 1 stays alone, another 1 stays alone, 3 stay together. (same as case3)
    case6*: 1 stay alone, 2 stay together, another 2 stay together. (same as case6)
    in fact, other cases are just repetitions of case 1 - 7.

    solve, how many ways in each cases, then add them.


    Quote Originally Posted by Misa-Campo View Post
    b) Suppose 2 travelers are husband and wife and must go to the same hotel. how many different accommodation arrangements are there if the other 3 can go to any of the other hotels?

    Please give me some tips for these questions, i dont have a clue on how to do these, thanks alot.
    among the cases i have listed, which will you add up?
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  5. #5
    Junior Member Misa-Campo's Avatar
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    Thanks alot!

    Wow, thank you so much for your time and support, very simplified and fast, i understand from this even better than my teachers explanation

    I understand the question on the car number, but im still a little confused on Q2) and Q3)



    Quote Originally Posted by kalagota View Post
    how many choices do you have in the hundreds digit?
    how many in the tens digit?
    in ones digit?

    take note you need a number less that 4000, and there are no repetitions..
    Ok for this one im srry i mistyped 400 but i was meant to write 4000
    so heres the question again

    How many numbers less than 4000 can be made using the digits 1,2,3,4,5,6,7 which are even numbers?

    3 choices for the thousands (need number less than 4 since it must be <4000)

    6 choices for the hundreds

    5 choices for the tens

    and finally 3 choices for the ones (since its must have have an even digit to be an even number???)

    oh and also the number 4000

    so i would have
    3x6x5x3+1 permutations in total which is 271 arrangements.

    Again thanks for your support on this question, however im still not sure if its right since these questions are tricky, can you check the answer for me if you can thanks alot.


    Quote Originally Posted by kalagota View Post
    case1: all five stay together in one hotel. how many ways?
    case2: 4 stay together on 1 hotel, the other 1 is isolated. how many ways?
    case3: 3 stay together on 1 hotel, 1 stays alone and also the other 1 stays alone.
    case4: 3 stay together on 1 hotel, 2 stays together on other hotel.
    case5: 2 stay together on 1 hotel, 3 distributes themselves to other hotels.
    case6: 2 stay together on 1 hotel, another 2 stay together on another hotel and 1 isolates.
    case4*: 2 stay together on 1 hotel, the other 3 stay on another hotel. (but this is the same as the case4. so this is not included anymore.)
    case7: all 5 distributes themselves without accompanying others. (each 1 stays alone on one hotel)
    case2*: 1 stays alone, 4 stay together. (same as case2.)
    case3*: 1 stays alone, another 1 stays alone, 3 stay together. (same as case3)
    case6*: 1 stay alone, 2 stay together, another 2 stay together. (same as case6)
    in fact, other cases are just repetitions of case 1 - 7.

    solve, how many ways in each cases, then add them.




    among the cases i have listed, which will you add up?
    This question i am still very, very confused, heres what i got:

    case1: all five stay together in one hotel. how many ways? 5!??
    case2: 4 stay together on 1 hotel, the other 1 is isolated. how many ways? 2! x 4!
    case3: 3 stay together on 1 hotel, 1 stays alone and also the other 1 stays alone. 3! x 3! x 1 x 1
    case4: 3 stay together on 1 hotel, 2 stays together on other hotel. 2! x 3! x 2!
    case5: 2 stay together on 1 hotel, 3 distributes themselves to other hotels.
    4! x 2! x 1 x 1 x 1
    case6: 2 stay together on 1 hotel, another 2 stay together on another hotel and 1 isolates.
    3! x 2! x 2! x 1
    case7: all 5 distributes themselves without accompanying others. (each 1 stays alone on one hotel) 5!

    when i add all this up i get 408 but the answer says 3125, where did i go wrong?

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  6. #6
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Misa-Campo View Post
    Wow, thank you so much for your time and support, very simplified and fast, i understand from this even better than my teachers explanation

    I understand the question on the car number, but im still a little confused on Q2) and Q3)

    Ok for this one im srry i mistyped 400 but i was meant to write 4000
    so heres the question again

    How many numbers less than 4000 can be made using the digits 1,2,3,4,5,6,7 which are even numbers?

    LOOK AT THE ORDER OF CHOOSING I WILL DO..


    a.. 3 choices for the thousands (need number less than 4 since it must be <4000)

    b.. and finally 3 choices for the ones (since its must have have an even digit to be an even number???)

    (however, a and b are dependent to each other.. since in a, if 2 were chosen, then there are only 2 choices in b)

    5 choices for the hundreds

    4 choices for the tens



    oh and also the number 4000

    so i would have
    3x6x5x3+1 permutations in total which is 271 arrangements.

    Again thanks for your support on this question, however im still not sure if its right since these questions are tricky, can you check the answer for me if you can thanks alot.
    that is, you have to choose for the ones digit first before the tens and the hundreds..
    Last edited by kalagota; July 12th 2008 at 08:08 PM.
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  7. #7
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Misa-Campo View Post
    This question i am still very, very confused, heres what i got:

    case1: all five stay together in one hotel. how many ways? 5!??
    case2: 4 stay together on 1 hotel, the other 1 is isolated. how many ways? 2! x 4!
    case3: 3 stay together on 1 hotel, 1 stays alone and also the other 1 stays alone. 3! x 3! x 1 x 1
    case4: 3 stay together on 1 hotel, 2 stays together on other hotel. 2! x 3! x 2!
    case5: 2 stay together on 1 hotel, 3 distributes themselves to other hotels.
    4! x 2! x 1 x 1 x 1
    case6: 2 stay together on 1 hotel, another 2 stay together on another hotel and 1 isolates.
    3! x 2! x 2! x 1
    case7: all 5 distributes themselves without accompanying others. (each 1 stays alone on one hotel) 5!

    when i add all this up i get 408 but the answer says 3125, where did i go wrong?

    for case1: consider the group as 1 entity? as one, how many choices does it have? there are only 5 hotels, and so, it has only 5 choices..

    case2: there are 2 entities right? the first entity can choose among the 5 hotels, and the other entity can have only 4 choices since the first has already chosen 1. so there are 5 x 4..


    do the rest with similar arguments..
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