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Permutation and Combination |
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| Aug27-08, 08:06 AM | #1 |
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Permutation and Combination
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? |
| Aug27-08, 09:04 AM | #2 |
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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??
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| Aug27-08, 11:43 AM | #3 |
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| Aug27-08, 11:47 AM | #4 |
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Permutation and Combination
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
regards |
| Aug27-08, 01:47 PM | #5 |
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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.
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| Aug28-08, 07:55 AM | #6 |
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