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From 7 consonants and 5 vowels, how many words can be formed [#permalink]

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28 Dec 2005, 19:45

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From 7 consonants and 5 vowels, how many words can be formed consisting of 4 different consonants and 3 different vowels? The words need not have meaning.

Re: PS - Probability Consonants & Vowels [#permalink]

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28 Dec 2005, 22:30

Itzrevs wrote:

From 7 consonants and 5 vowels, how many words can be formed consisting of 4 different consonants and 3 different vowels? The words need not have meaning.

Please explain your answer!!!!

Number of ways to choose 4 cons out of 7*Number of ways to choose 3 vowels out of 5---->4C7*3C5=(7*6*5*4/1*2*3*4)*(5*4*3/1*2*3)=350

Re: PS - Probability Consonants & Vowels [#permalink]

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29 Dec 2005, 01:18

Itzrevs wrote:

From 7 consonants and 5 vowels, how many words can be formed consisting of 4 different consonants and 3 different vowels? The words need not have meaning.

Please explain your answer!!!!

Since the order of the letters are important we get
7P4*5P3 = 4200

Re: PS - Probability Consonants & Vowels [#permalink]

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29 Dec 2005, 10:54

krisrini wrote:

Itzrevs wrote:

From 7 consonants and 5 vowels, how many words can be formed consisting of 4 different consonants and 3 different vowels? The words need not have meaning.

Please explain your answer!!!!

Since the order of the letters are important we get 7P4*5P3 = 4200

u have put the right formula but calculated it incorrectly. ur answer shud also be 50400....
_________________

Re: PS - Probability Consonants & Vowels [#permalink]

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29 Dec 2005, 22:22

crazyfin wrote:

krisrini wrote:

Itzrevs wrote:

From 7 consonants and 5 vowels, how many words can be formed consisting of 4 different consonants and 3 different vowels? The words need not have meaning.

Please explain your answer!!!!

Since the order of the letters are important we get 7P4*5P3 = 4200

u have put the right formula but calculated it incorrectly. ur answer shud also be 50400....

Thanks crazyfin. The compuatation should have worked to 50400, I am not sure how I made this mistake, thanks much for pointing out.

The 4 different consonants can be selected in 7C4 ways
The different vowels can be selected in 5C3 ways
The resulting 7 different letters ( 4 consonants & 3 vowels) can be arranged themselves in 7P7 = 7! ways..

So the answer is 7C4 * 5C3 * 7! = 35 *10 * 5040 = 1764000
_________________

"The difference between a smart person and a wise person is that
a smart person knows what to say and
a wise person knows whether or not to say it."

The 4 different consonants can be selected in 7C4 ways The different vowels can be selected in 5C3 ways The resulting 7 different letters ( 4 consonants & 3 vowels) can be arranged themselves in 7P7 = 7! ways..

So the answer is 7C4 * 5C3 * 7! = 35 *10 * 5040 = 1764000

Re: From 7 consonants and 5 vowels, how many words can be formed [#permalink]

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05 Nov 2013, 19:47

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Re: From 7 consonants and 5 vowels, how many words can be formed [#permalink]

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23 Nov 2013, 06:12

Itzrevs wrote:

From 7 consonants and 5 vowels, how many words can be formed consisting of 4 different consonants and 3 different vowels? The words need not have meaning.

Because each word has to consist of different consonants or vowels, the correct answer is (7*6*5*4)*(5*4*3) and not 7^4*5^3. I used bars to distinguish the numbers representing consonants and vowels to make it more clear.

From 7 consonants and 5 vowels, how many words can be formed consisting of 4 different consonants and 3 different vowels? The words need not have meaning.

That's because we are arranging all 7 letters together.

For example, consider simpler case: how many 4-letter words you can make from 2 consonants {b, c} and two vowels {a, e}. If you arrange them first ({b, c} and {c, b} for consonants and {a, e} and {e, a} for vowels) you'll get 2*2=4 words, which would be wrong. The correct answer is 4!=24.
_________________

Are there ways to come up with sub-types of this particular question?

For example,

A case with 3 distinct vowels & 4 distinct consonants. A case where the ORDER of the letters DOES NOT matter (OR) a case where ORDER matters. etc.

OR is there a particular link where such questions are discussed?

Thanks in advance.

The question at hand is where we have distinct letters and the order does matters. If the order does not matter the answer would be \(C^4_7*C^3_5=35*10=350\).

Re: From 7 consonants and 5 vowels, how many words can be formed [#permalink]

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02 May 2014, 09:51

Hi Bunuel What is wrong with my approach. Can you please correct me : Imagine 7 boxes now let's say we are filling first 4 boxes with consonants .First box can be filled with 7 ways.Second with 6 and so on. Similarly for 3 vowel boxes : first box can be formed with 5 ways , second with 4 ways and third with 3 ways :

So total number of ways =7*6*5*4*5*4*3 And now these letters can be arranged in 7! ways to have different combination. I know I am wrong. Can you please correct me.

Thanks in Advance.
_________________

Feel Free to Press Kudos if you like the way I think .

Hi Bunuel What is wrong with my approach. Can you please correct me : Imagine 7 boxes now let's say we are filling first 4 boxes with consonants .First box can be filled with 7 ways.Second with 6 and so on. Similarly for 3 vowel boxes : first box can be formed with 5 ways , second with 4 ways and third with 3 ways :

So total number of ways =7*6*5*4*5*4*3 And now these letters can be arranged in 7! ways to have different combination. I know I am wrong. Can you please correct me.

Thanks in Advance.

Short answer would be that 7*6*5*4 as well as 5*4*3 will contain repetitions. The number of way to choose 4 letters out of 7 is NOT 7*6*5*4 it's \(C^4_7=\frac{7!}{4!3!}=35\) and the number of way to choose 3 letters out of 5 is NOT 5*4*3 it's \(C^3_5=\frac{5!}{3!2!}=10\).
_________________

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