How many 5-letter words can be formed using the alphabet of English language that contain 2 different vowels and 3 different consonants?
A. 5C2 * 21C3
B. 5P2 * 21P3
C. 5C2 * 21C3 * 5!
D. 5P2 * 21P3 * 5!
E. 52 * 213 * 5!
Not sure what the answer is. I think it should be B.
2 vowels and 3 consonants
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- cypherskull
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I think the answer is (D).
From 21 consonants and 5 vowels.
# of ways to choose 3 different consonants: 21 * 20 * 19 = P(21,3)
# of ways to choose 2 different vowels: 5 * 4 = P(5,2)
Since order counts (producing different words), there are 5! ways to arrange the letters once they are selected.
From 21 consonants and 5 vowels.
# of ways to choose 3 different consonants: 21 * 20 * 19 = P(21,3)
# of ways to choose 2 different vowels: 5 * 4 = P(5,2)
Since order counts (producing different words), there are 5! ways to arrange the letters once they are selected.
I'm really old, but I'll never be too old to become more educated.
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Isn't Permutation an ordered combination in itself? I chose B because I think the formula for permutation itself takes into account the order of the selection. Please correct me if I'm wrong.
Thanks,
Sunit
Thanks,
Sunit
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Here's how I approached it...we need to select 3 different consonants and 2 different vowels.
Since there are 21 consonants in the alphabet, there are 21 ways to select the first of three consonants. The next consonant must be different, so there are now 20 ways to select the 2nd consonant. The third consonant must be different from the first two selections, so there are 19 ways to select the third consonant.
Therefore, there are 21*20*19 (or P(21,3)) ways of selecting the three consonants.
Since there are 5 vowels in the alphabet, there are 5 ways to select the first of two vowels to be used. The second vowel must be different from the first vowel, so there are 4 ways to select the second vowel.
Therefore, there are 5*4 (or P(5,2)) ways of selecting the two vowels.
Therefore, there are P(21,3) * P(5,2) ways of selecting the 5 letters to be used for our words.
If the question asked only to calculate the number of ways that 3 different consonants and 2 different vowels can be selected from the alphabet, then we'd be done. However, the question asks how many words can be formed, and there are 5! ways to arrange the letters once they are selected.
Then again, maybe I'm just reading the question wrong.
Since there are 21 consonants in the alphabet, there are 21 ways to select the first of three consonants. The next consonant must be different, so there are now 20 ways to select the 2nd consonant. The third consonant must be different from the first two selections, so there are 19 ways to select the third consonant.
Therefore, there are 21*20*19 (or P(21,3)) ways of selecting the three consonants.
Since there are 5 vowels in the alphabet, there are 5 ways to select the first of two vowels to be used. The second vowel must be different from the first vowel, so there are 4 ways to select the second vowel.
Therefore, there are 5*4 (or P(5,2)) ways of selecting the two vowels.
Therefore, there are P(21,3) * P(5,2) ways of selecting the 5 letters to be used for our words.
If the question asked only to calculate the number of ways that 3 different consonants and 2 different vowels can be selected from the alphabet, then we'd be done. However, the question asks how many words can be formed, and there are 5! ways to arrange the letters once they are selected.
Then again, maybe I'm just reading the question wrong.
I'm really old, but I'll never be too old to become more educated.
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The answer is obviously c.
21c3 * 5*2 * 5!
order does not matter, therefore combination
21 consonants to choose from to fill the 3 spots, and 5 vowels to choose from to fill the rest of the 2 spots. Once you get the combo of possibilities, you need to then rearrange the letters and in this case ORDER matters so 5P5 aka 5!. Hence the answer is c)
PS this is like grade 11 math cmon guys
21c3 * 5*2 * 5!
order does not matter, therefore combination
21 consonants to choose from to fill the 3 spots, and 5 vowels to choose from to fill the rest of the 2 spots. Once you get the combo of possibilities, you need to then rearrange the letters and in this case ORDER matters so 5P5 aka 5!. Hence the answer is c)
PS this is like grade 11 math cmon guys
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cypherskull wrote:How many 5-letter words can be formed using the alphabet of English language that contain 2 different vowels and 3 different consonants?
A. 5C2 * 21C3
B. 5P2 * 21P3
C. 5C2 * 21C3 * 5!
D. 5P2 * 21P3 * 5!
E. 52 * 213 * 5!
Not sure what the answer is. I think it should be B.
I hope this wouldn't be a real question. The question asks for "words". I am not aware of an English word that has three consonants in a row followed by two vowels.
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The problem as written would not appear on the GMAT.regor60 wrote:I hope this wouldn't be a real question. The question asks for "words". I am not aware of an English word that has three consonants in a row followed by two vowels.
For one thing, the GMAT does not use notation such as 5C2 or 5P2.
That said, it is possible to form a valid English word using different three consonants in a row followed by two different vowels (stria, for example).
Last edited by GMATGuruNY on Mon Mar 09, 2015 6:01 pm, edited 1 time in total.
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As Mitch points out, the GMAT would never word a question this way. That said, here's one way to answer the question:cypherskull wrote:How many 5-letter words can be formed using the alphabet of English language that contain 2 different vowels and 3 different consonants?
A. 5C2 * 21C3
B. 5P2 * 21P3
C. 5C2 * 21C3 * 5!
D. 5P2 * 21P3 * 5!
E. 52 * 213 * 5!
Take the task of creating 5-letter words and break it into stages.
Stage 1: Select the 2 vowels you'll be using
Since the order in which we SELECT the vowels does not matter, we can use combinations.
We can select 2 vowels from 5 vowels in 5C2 ways (= 10 ways)
So, we can complete stage 1 in 10 ways
If anyone is interested, we have a free video on calculating combinations (like 5C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789
Stage 2: Select the 3 consonants you'll be using
Since the order in which we SELECT the consonants does not matter, we can use combinations.
We can select 3 consonants from 21 vowels in 21C3 ways (= 1330 ways)
So, we can complete stage 2 in 1330 ways
NOTE: Now that we've selected the 5 letters, it's time to arrange them.
Stage 3: Arrange the 5 letters in a row.
This can be accomplished in 5! ways (120 ways)
So, we can complete stage 3 in 120 ways
By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create a 5-letter word) in (10)(1330)(120) ways
ORRR, we can say that all 3 stages can be completed in [spoiler](5C2)(21C3)(5!)[/spoiler] ways
Answer: C
--------------------------
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First, most of the math tested on the GMAT doesn't go beyond grade 11. So, your comment is somewhat moot since it applies to almost all GMAT quant questions.BetterThanU wrote: PS this is like grade 11 math cmon guys
Second, the difficulty level of this question is probably in 600-700 range.
Third, this forum runs better if we refrain from chiding others.
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Agreed!GMATGuruNY wrote: The problem as written would not appear on the GMAT.
For one thing, the GMAT does not use notation such as 5C2 or 52P.
One reason for this is that it's often possible to solve counting questions using techniques other than applying the rules for combinations and permutations. So, giving answers with this notation would be too restrictive.
Another reason is that counting notation is not universal. For example, while some people might use the notation 5C2 to represent the idea of "5 choose 2," others use bracket notation as shown below.
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Totally agree. The content required is actually junior high level, though the applications are significantly more difficult. (The level of the content is misleading: for example, the USAMO is "high school level", but the applications are difficult enough that the best students in the country take the test yet the median score is 4 out of 42.)Brent@GMATPrepNow wrote:First, most of the math tested on the GMAT doesn't go beyond grade 11. So, your comment is somewhat moot since it applies to almost all GMAT quant questions.
The GMAT is mostly about being very proficient in the basics - which few people are - and not missing tricky/nitpicky details under time constraints and test day stress: more than anything else, in my opinion, it tests the acuity of your _first instinct_. (If you don't have a good idea that you can build on almost immediately, the clock will get you.) So the questions are less deep insight and more pattern recognition and attention to detail.
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Definitely!Matt@VeritasPrep wrote:Totally agree. The content required is actually junior high level, though the applications are significantly more difficult. (The level of the content is misleading: for example, the USAMO is "high school level", but the applications are difficult enough that the best students in the country take the test yet the median score is 4 out of 42.)Brent@GMATPrepNow wrote:First, most of the math tested on the GMAT doesn't go beyond grade 11. So, your comment is somewhat moot since it applies to almost all GMAT quant questions.
The GMAT is mostly about being very proficient in the basics - which few people are - and not missing tricky/nitpicky details under time constraints and test day stress: more than anything else, in my opinion, it tests the acuity of your _first instinct_. (If you don't have a good idea that you can build on almost immediately, the clock will get you.) So the questions are less deep insight and more pattern recognition and attention to detail.
For example, although the concept of EVEN and ODD integers is taught in elementary school, some of the GMAT questions that test this concept would many seasoned math teachers.
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Brent