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From 7 consonants and 5 vowels, how many words can be formed

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Itzrevs
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From 7 consonants and 5 vowels, how many words can be formed [#permalink] New post   Updated on: 23 Nov 2013, 06:18
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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.

OA:
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1,764,000

Originally posted by Itzrevs on 28 Dec 2005, 19:45.
Last edited by Bunuel on 23 Nov 2013, 06:18, edited 1 time in total.
Added the OA.
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Re: From 7 consonants and 5 vowels, how many words can be formed [#permalink] New post   23 Nov 2013, 06:22
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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.

OA:
Show Spoiler
1,764,000


The # of ways to choose 4 different consonants out of 7 is \(C^4_7=35\);
The # of ways to choose 3 different vowels out of 5 is \(C^3_5=10\).

Thus the # of ways to choose 4 different consonants and 3 different vowels is \(35*10=350\).

Finally these 7 letters can be arranged in 7! ways, therefore the total # of different words is \(350*7!=1,764,000\).

Hope this helps.
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Yurik79
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Re: PS - Probability Consonants & Vowels [#permalink] New post   28 Dec 2005, 22:30
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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
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Re: PS - Probability Consonants & Vowels [#permalink] New post   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
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[#permalink] New post   29 Dec 2005, 08:05
sorry guys both of you didn't get the OA..

Lets give it another try ;-)
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[#permalink] New post   29 Dec 2005, 08:55
4 consonants can be chosen in 7*6*5*4 ways

3 vowels can be chosen in 5*4*3 ways

Total no. of alternatives =7*6*5*4*5*4*3=50,400
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Re: PS - Probability Consonants & Vowels [#permalink] New post   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....
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Itzrevs
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[#permalink] New post   29 Dec 2005, 21:50
Yurik79, I didn't the same way as you did. But thats not the answer.

I shall wait for few more people to jump in and post their answers & explanations..

Will post the OA soon.
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Re: PS - Probability Consonants & Vowels [#permalink] New post   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.
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[#permalink] New post   30 Dec 2005, 10:17
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Guys, The OA is 1,764,000

OE :

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
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[#permalink] New post   30 Dec 2005, 12:33
Itzrevs wrote:
Guys, The OA is 1,764,000

OE :

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

It seemed so simple at first look)))Good one!10x
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Re: From 7 consonants and 5 vowels, how many words can be formed [#permalink] New post   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.
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Re: From 7 consonants and 5 vowels, how many words can be formed [#permalink] New post   25 Nov 2013, 23:13
I do not think you will see like this question in the test, unless the answer choices contain 350 * 7!.

but this is a good question to practise both concept - combination and permutation- in one question.
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Re: From 7 consonants and 5 vowels, how many words can be formed [#permalink] New post   01 May 2014, 09:05
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Hi,
I still don't get whats wrong with 7P4*5P3? :?
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Re: From 7 consonants and 5 vowels, how many words can be formed [#permalink] New post   01 May 2014, 09:53
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shashankprasad wrote:
Hi,
I still don't get whats wrong with 7P4*5P3? :?


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.
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Re: From 7 consonants and 5 vowels, how many words can be formed [#permalink] New post   01 May 2014, 22:37
Hello there

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.
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Re: From 7 consonants and 5 vowels, how many words can be formed [#permalink] New post   02 May 2014, 02:21
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shaderon wrote:
Hello there

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 \(C^4_7*C^3_5=35*10=350\).

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html


Hope this helps.
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Re: From 7 consonants and 5 vowels, how many words can be formed [#permalink] New post   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.
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Re: From 7 consonants and 5 vowels, how many words can be formed [#permalink] New post   02 May 2014, 10:13
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282552 wrote:
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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Re: From 7 consonants and 5 vowels, how many words can be formed [#permalink] New post   23 Apr 2024, 04:07
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Re: From 7 consonants and 5 vowels, how many words can be formed [#permalink]
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