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Permutations

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by shankar.ashwin » Wed Sep 07, 2011 6:02 am
How many 6 letter word can be formed such that there are 2 vowels and 4 consonants in it when
(1) Repetition is not allowed
(2) Repetition is allowed.

I have seen a similar sum in the forum, but could anyone help me out in the case when repetition is allowed?
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by cans » Wed Sep 07, 2011 6:49 am
a) 5C2*21C4*6! (select 2 vowels and 4 consonants and arrange them)
b) case 1: no repetition: 5C2*21C4*6!
case 2: only vowels repeated: 5C1*21C4*(6!/2!)
case 3: 2 consonants repeated (bbcd...): 5C2*21C3*(6!/2!)
case 4: 3 consonants repeated (bbbc): 5C2*21C2*6!/3!
case 5: 4 consonants (bbbb): 5C2*21C1*(6!/4!)
case 6: (bbcc): 5C2*21C2*(6!/2!/2!)
case 7: combo of case 2 and case 3. - 5C1*21C3*(6!/2!/2!)
case 8: case 2 + case 4
case 9: case 2+ case 5
case 10: case 2 + case 6
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by shankar.ashwin » Wed Sep 07, 2011 7:37 am
Wow, thanks. I couldnt comprehend all the cases, but I get what you are doing. Unlikely to see a sum like this. But helpful to get your concepts right. Thanks
cans wrote:a) 5C2*21C4*6! (select 2 vowels and 4 consonants and arrange them)
b) case 1: no repetition: 5C2*21C4*6!
case 2: only vowels repeated: 5C1*21C4*(6!/2!)
case 3: 2 consonants repeated (bbcd...): 5C2*21C3*(6!/2!)
case 4: 3 consonants repeated (bbbc): 5C2*21C2*6!/3!
case 5: 4 consonants (bbbb): 5C2*21C1*(6!/4!)
case 6: (bbcc): 5C2*21C2*(6!/2!/2!)
case 7: combo of case 2 and case 3. - 5C1*21C3*(6!/2!/2!)
case 8: case 2 + case 4
case 9: case 2+ case 5
case 10: case 2 + case 6
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