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permutation problem need help

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by agoyal2 » Wed Jun 03, 2009 6:03 am
We have to arrange 3 consonants and 2 vowels.

So, one possible arrangement is CCCVV
All possible arrangements = 5!/ (2!3!) = 10

Now, we have to choose 3 consonants from available 4 = 4P3
Choose 2 vowels from available 3 = 3P2

Total = 4P3 * 3P2 * 10 = 1440

I got a 144 earlier but looking at the OA, I re-worked and came to the above calculation. Will have to be careful of these kind of problems. Thanks for posting!
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by crejoc » Wed Jun 03, 2009 6:57 am
I too initially came up with the answer 144, thanks for the explanation..
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by Musiq » Wed Jun 03, 2009 10:06 am
agoyal2 wrote:We have to arrange 3 consonants and 2 vowels.

So, one possible arrangement is CCCVV
All possible arrangements = 5!/ (2!3!) = 10
I dont get this part...how does the Combination formula come into play here?

Could you please explain...Thanks!
For love, not money.
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by crejoc » Wed Jun 03, 2009 11:00 am
Musiq wrote:
agoyal2 wrote:We have to arrange 3 consonants and 2 vowels.

So, one possible arrangement is CCCVV
All possible arrangements = 5!/ (2!3!) = 10
I dont get this part...how does the Combination formula come into play here?

Could you please explain...Thanks!
Musiq
Actually it is not combination formula..
it also permutation, that is the arrangements of the selected vowels and consonants among themselves. as we are selecting 5 letters , the permutation has to be 5! , but two are of same type (vowels ) and other three of other type(consonants) so dividing by (2!*3!). hope am right..
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by Musiq » Wed Jun 03, 2009 12:00 pm
crejoc wrote:
Musiq wrote:
agoyal2 wrote:We have to arrange 3 consonants and 2 vowels.

So, one possible arrangement is CCCVV
All possible arrangements = 5!/ (2!3!) = 10
I dont get this part...how does the Combination formula come into play here?

Could you please explain...Thanks!
Musiq
Actually it is not combination formula..
it also permutation, that is the arrangements of the selected vowels and consonants among themselves. as we are selecting 5 letters , the permutation has to be 5! , but two are of same type (vowels ) and other three of other type(consonants) so dividing by (2!*3!). hope am right..

Aaaaah......makes sense to me now.

I understood it from the Permutations perspective....but the number of arrangements looked like 5C3 (the way it was written).

Thanks for the explanation.
For love, not money.
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by ghacker » Tue Jun 09, 2009 6:16 am
PROMISE = 3V and 4C

all the letters are different so , we can select 2V is 3 ways and 3C is 4 ways

but can arrange the selected letters in 5! ways

Total no of ways = 3*4*5! = 12*120 = 1440
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Re: permutation problem need help

by Stuart@KaplanGMAT » Tue Jun 09, 2009 2:55 pm
crejoc wrote:How many words (arrangements of letters) containing 3 consonants and 2 vowels can be formed from the letters of the word promise?

kindly explain..
First, we need to choose our letters.

For consonants, we want to select 3 out of 4. For vowels, we want to select 2 out of 3.

We don't care about the order in which we select them, so we use the combinations formula. We're selecting 3 consonants AND 2 vowels, so we MULTIPLY the results:

4C3 * 3C2 = 4!/3!1! * 3!/2!1! = 4*3 = 12

Once we select our letters, we need to arrange them into different configurations. We have 5 letters to arrange with no duplicates (the answer would be different if we had two "E"s on our list, for example). Since we're arranging we care about order, so we use the permutations formula:

5P5 = 5!/0! = 5! = 5*4*3*2 = 120

So, we have 12 different selections of letters and 120 ways to arrange our letters once selected. We're doing multiple steps, so we MULTIPLY the results:

12*120 = 1440.

Some important formulas:

nCk = n!/k!(n-k)!

nPk = n!/(n-k)!

in both formulas, n=total # of objects and k=# of objects used.

For permutations, if n=k (i.e. you're using all the objects), the total number of arrangements is simply n!.

Some good combinations shortcuts to remember:

nCn = 1
nC0 = 1
nC1 = n
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by rah_pandey » Tue Jun 09, 2009 10:44 pm
I think the last explaination is the correct approach. CCCVV are all diff there fore no of words possible = 5!.
and ways of choosing 3 C and 2 V =12

we should not try to find permutation while selecting vowels or consonants. This means we are arranging the consonants among themselves but they have to be arranged in the entire word.

Regards,
Rahul
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by tohellandback » Tue Jun 09, 2009 10:56 pm
agoyal,
The consonants and the vowels are not similar. They are different letters.
so i don't think 5!/2!3! holds true

Thanks
The powers of two are bloody impolite!!
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