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!
permutation problem need help
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Musiq
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I dont get this part...how does the Combination formula come into play here?agoyal2 wrote:We have to arrange 3 consonants and 2 vowels.
So, one possible arrangement is CCCVV
All possible arrangements = 5!/ (2!3!) = 10
Could you please explain...Thanks!
For love, not money.
MusiqMusiq wrote:I dont get this part...how does the Combination formula come into play here?agoyal2 wrote:We have to arrange 3 consonants and 2 vowels.
So, one possible arrangement is CCCVV
All possible arrangements = 5!/ (2!3!) = 10
Could you please explain...Thanks!
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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Musiq
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crejoc wrote:MusiqMusiq wrote:I dont get this part...how does the Combination formula come into play here?agoyal2 wrote:We have to arrange 3 consonants and 2 vowels.
So, one possible arrangement is CCCVV
All possible arrangements = 5!/ (2!3!) = 10
Could you please explain...Thanks!
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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ghacker
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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
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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Stuart@KaplanGMAT
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First, we need to choose our letters.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..
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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rah_pandey
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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
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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tohellandback
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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 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!!
