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Permutation and Combination

  1. Aug 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Consider the word MATHEMATICS. There are some vowels: AEAI
    The remaining 7 letters are MTHMTCS. Find the number of different 11 lettered words formed from these particular letters (repetition not allowed) such that all the vowels occur in the same order AEAI.
    For example:
    SCAHTEAIMMT



    3. The attempt at a solution
    The total number of words that can be formed is 11! without any restrictions.
    If we fix the A at place 1:
    no. of combinations of 3 from remaining: 10C3
    If we fix the A at place 2:
    no. of combinations of 3 from remaining: 9C3
    If we fix the A at place 3:
    no. of combinations of 3 from remaining: 8C3
    .....
    .....
    Therefore the number of arrangements should be: 10C3+9C3+8C3+.........+3C3
    =330
    which gives me a wrong answer. Why? What should I do to get a correct answer?
     
  2. jcsd
  3. Aug 27, 2008 #2

    Dick

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    You aren't accounting for the ways to place the consonants after the vowels are fixed. You are also counting the vowel placements the hard way. Why not just 11C4? Who cares where 'A' is??
     
  4. Aug 27, 2008 #3
    Ok, I admit I am wrong. I am completely stuck with the question. Please help me out!!!!!!!!
     
  5. Aug 27, 2008 #4
    Its seems a bit hard to me. Nothing is fixed except the order of the vowels. What shall I do to this. Assuming seven places betwen each vowel doesnt help. I have also tried to fix the consonants first. Help :smile:
    regards
     
    Last edited: Aug 27, 2008
  6. Aug 27, 2008 #5

    Dick

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    You AREN'T wrong so far. You just aren't finished. You counted the vowel placements correctly (though as, I say, the hard way). So take one of your vowel arrangements. There are seven empty spaces left and you have seven consonants to put in them. How many ways can you do that?? The number of consonant arrangements doesn't depend on the particular vowel arrangement, right? So you can just multiply them.
     
  7. Aug 28, 2008 #6
    Yeah, I have realised that by now. Thanks a lot for ur help :smile:
     
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