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Homework Help: Permutation and combination in maths

  1. Sep 12, 2005 #1
    Hi. I've got problem with this task.
    We have 5 digit. How many 7-digit numbers can we create that has at least 2 different digit?

  2. jcsd
  3. Sep 12, 2005 #2
    whats the problem?
  4. Sep 12, 2005 #3
    I don't know how to solve the task.

    Could anybody help me?
    Last edited: Sep 13, 2005
  5. Sep 14, 2005 #4
    I think that the answer is 5^6=15625 but i don't know how to prove it. Could you help me?
  6. Sep 14, 2005 #5


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    Homework Helper

    I think this problem is related to permutation and combination in maths.
  7. Sep 14, 2005 #6


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    There could be some confusion about what the question actually asks. If you stated the question exactly as given that might help.

    As I understand it, you are ask to create a seven-digit number where each digit comes from a set of five choices, and there must be at least two different digits represented within the seven. If that is the right interpretation of the question, there are a number of ways of going about this.

    The secret in combinatorics is to look for the short cuts to the answer. There are usually different equivalent ways of enumerating things. In this case, the quick way to the answer is to count up the options you're not allowed to have.

    Think of it this way. Every possible combination of digits is allowed (each of seven digits with 5 choices) except for a certain set. If you can figure what is the set that isn't allowed, you can just subtract it from the total.

    I won't give away the next step but ask yourself what combinations aren't allowed and then work out how many of them there are.
  8. Sep 14, 2005 #7
    I hope this is the correct answer to your question :

    (5 ) (5) * 5 ! - 5 =
    (25) ( 120) - 5 = 2995 ways
    Last edited: Sep 14, 2005
  9. Sep 14, 2005 #8
    Why is it correct answer? Could you explain it? I think it is too small.
  10. Sep 14, 2005 #9
    Yes it is P&C problem;
    well first off given only 5 distinct digits(i assume thats what you meant and lets call it alphabet to distinguish from 7digits)
    [0]how many 7 digit numbers can you make with 5 digits(- - - - - - -) =
    that is the total you can have. REMEMBER TO GET RID OF REPETITIONS
    find an easy way to do this..hint if (a is the first digit then a must also be the last.

    [1][Now what is the exclusion(well you want at least 2 different alphabets): you don't want any 7digit that has exactly only 1 alphabet present...
    thus your exclusion set is #of 7digits of only 1 alphabet present...THUS YOU GOTTA FIGURE HOW MANY THERE ARE

    Your solution will be #=TOTAL-Exlusion set.

    lastly becareful how many times your exclusion set appears in the set of repetitions.
    Last edited: Sep 14, 2005
  11. Sep 14, 2005 #10
    I think I made small mistake , you have to choose five choices of each digit of each of the seven digits you suppose to fill then you have


    excluding five choices (a,a,a,a,a) or (b,b,b,b,b) .........five of them
    then the answer = 5^7-5=78120 ways

    I am sorry again I made mistake.I hope this time it is right
  12. Sep 14, 2005 #11
    If you mean you have to make a 5-number sequence when you have 7 different numbers to choose from for each position, you need to do (number of possibilities per position)^(number of positions). Subtract the illegal choices from that number, and you're done.
    Last edited: Sep 14, 2005
  13. Sep 14, 2005 #12
    mithal i dont' think thats the right answer because you forgot to subtract repetitive #s from the 5^7.
  14. Sep 14, 2005 #13
    I think it is right because these five choices are the only non - permissable in the condition of the problem ( at least two allowed different digits ) suppose else (a,a,a,a,b) is included a,b are two different digits .
  15. Sep 14, 2005 #14
    thats not what i meant.... what i meant was the set of all valid 7digits prior to the restriction (subtraction of 5) is not 5^7.
  16. Sep 14, 2005 #15
    It is 5 ^7 .Every time you have to make five choices for each digit of the seven digits .repetion is allowed .So in this way you cover all possibilities and then subtract the five cases that not allowed in the condition of the problem as I mentioned before .
  17. Sep 14, 2005 #16
    No, repetitions are not allowed. repetitions are not distinct
    for examples in 3digit words 121 and 121 are a repetition but in 5^3 they are counted separatedly so 2x rather than 1x.
  18. Sep 14, 2005 #17
    I didn't mean repeation in the sense you are talking about . I mean if you choose 3 for the first digit then you can use 3 again for the second digit or the third or etc. or any other number . That mean once you choose that number for that digit you can repeat it again subsequently for other digits.You can visualize this if you draw a tree and consider different cases .

    I also would like to point out to a small mistake I made with the notations

    I assumed the five digits are (a , b , c ,d ,e ) and

    I meant ( a , a ,a ,a ,a ,a ,a ) and not like in the solution
    ( a , a ,a ,a ,a ) since it is seven digit number not five and the same for the others . But this has no effects on the logic of the answer . All stay the same .
  19. Sep 14, 2005 #18
    i understood what repetition you were talking about. However i was not refereing to the alphabet(5digits) but rather the word(7digit word)
    THe repetition i was thinking of was an undirectional ordering question where
    abc is the same cba. I realized i was wrong. and i apologize.
    The question was refering to directional words.
    Last edited: Sep 14, 2005
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