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Using the standard 26-letter alphabet, how many different groups of 4
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26 Apr 2017, 07:40
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26% (02:18) correct 74% (02:51) wrong based on 100 sessions
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Using the standard 26-letter alphabet, how many different groups of 4 letters are possible, if exactly two letters are chosen from a, e, i, o ,u and two letters are chosen from all the remaing letters except for y, and a letter can be used more than once?
Using the standard 26-letter alphabet, how many different groups of 4
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Updated on: 29 Apr 2017, 12:13
it's merely a selection problem: no. of different ways of selecting 4 letters as per question would be (5c2+5)*(20c2+20)=15*210=3150-> option D ; I have consider the repetitive part and the non-repetitive part of each type i.e vowels and consonant type and the total no. of selections is the product of the no. of selections of these two types.
Originally posted by attari92 on 26 Apr 2017, 08:05.
Last edited by attari92 on 29 Apr 2017, 12:13, edited 1 time in total.
Re: Using the standard 26-letter alphabet, how many different groups of 4
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26 Apr 2017, 09:18
I am not sure what i am doing wrong. 5 ways to choose a vowel. 2 vowels allowed -> 5*5 = 25 21 ways to choose a consonant. y cannot be chosen. 20 ways to choose. 2 consonants allowed -> 20*20.
Re: Using the standard 26-letter alphabet, how many different groups of 4
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27 Apr 2017, 21:26
mehrotrayashraj wrote:
I am not sure what i am doing wrong. 5 ways to choose a vowel. 2 vowels allowed -> 5*5 = 25 21 ways to choose a consonant. y cannot be chosen. 20 ways to choose. 2 consonants allowed -> 20*20.
Brining it all together -> 5*5*20*20 = 10,000
mehrotrayashraj your are conisdering "ae and ea" to be different.. the question stem say, different groups,, in group of "a,e, p,m" is same as "e,a,p,m".
in short u have used permutations,, which is not required here
Re: Using the standard 26-letter alphabet, how many different groups of 4
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29 Apr 2017, 11:36
attari92 wrote:
it's merely a selection problem: no. of different ways of selecting 4 letters as per question would be (10c2+5)*(20c2+20)=15*210=3150-> option D ; I have consider the repetitive part and the non-repetitive part of each type i.e vowels and consonant type and the total no. of selections is the product of the no. of selections of these two types.
Re: Using the standard 26-letter alphabet, how many different groups of 4
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29 Apr 2017, 12:16
skaur wrote:
attari92 wrote:
it's merely a selection problem: no. of different ways of selecting 4 letters as per question would be (10c2+5)*(20c2+20)=15*210=3150-> option D ; I have consider the repetitive part and the non-repetitive part of each type i.e vowels and consonant type and the total no. of selections is the product of the no. of selections of these two types.
I think you mean 5c2
phew! be to that typo yar but rest of the calculation was correct.Any way i do stand corrected
Re: Using the standard 26-letter alphabet, how many different groups of 4
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02 May 2017, 15:55
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Bunuel wrote:
Using the standard 26-letter alphabet, how many different groups of 4 letters are possible, if exactly two letters are chosen from a, e, i, o ,u and two letters are chosen from all the remaing letters except for y, and a letter can be used more than once?
A. 500 B. 1900 C. 2500 D. 3150 E. 6300
Since the order doesn’t matter and the letter can be repeated, the number of ways to choose 2 vowels is 5C2 + 5C1 = 10 + 5 = 15. (Notice that 5C2 is the number of ways to choose 2 vowels if they are different and 5C1 is the number of ways to choose 2 vowels if they are the same.)
Similarly, the number of ways to choose 2 non-y consonants is 20C2 + 20C1 = 190 + 20 = 210.
Thus, the number of ways to choose 2 vowels and 2 non-y consonants is 15 x 210 = 3150.
Answer: D
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Re: Using the standard 26-letter alphabet, how many different groups of 4
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02 May 2017, 22:49
Using the standard 26-letter alphabet, how many different groups of 4 letters are possible, if exactly two letters are chosen from a, e, i, o ,u and two letters are chosen from all the remaing letters except for y, and a letter can be used more than once?
A. 500 B. 1900 C. 2500 D. 3150 E. 6300
Note -As it says group assuming that the arrangement doesn't matter
group with both repetitive consonants and vowels = 4C1*20C1 = 80 group with repetitive consonants and non repetitive vowels = 20C1* 5C2 = 200 group with repetitive vowels and non repetitive consonants = 21C2* 4C1 = 840 group with both non repetitive consonants and vowels = 21C2*5C2 = 2100 total = 3220 Can anyone pls tell me where I went wrong
Re: Using the standard 26-letter alphabet, how many different groups of 4
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02 May 2017, 23:36
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sreenu7464 wrote:
Using the standard 26-letter alphabet, how many different groups of 4 letters are possible, if exactly two letters are chosen from a, e, i, o ,u and two letters are chosen from all the remaing letters except for y, and a letter can be used more than once?
A. 500 B. 1900 C. 2500 D. 3150 E. 6300
Note -As it says group assuming that the arrangement doesn't matter
group with both repetitive consonants and vowels = 4C1*20C1 = 80 ==> should be 5C1*20C1 group with repetitive consonants and non repetitive vowels = 20C1* 5C2 = 200 group with repetitive vowels and non repetitive consonants = 21C2* 4C1 = 840 ==> should be 20C2* 5C1 group with both non repetitive consonants and vowels = 21C2*5C2 = 2100 ==> should be 20C2*5C2 total = 3220 Can anyone pls tell me where I went wrong
Re: Using the standard 26-letter alphabet, how many different groups of 4
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04 Sep 2018, 03:17
ScottTargetTestPrep wrote:
Bunuel wrote:
Using the standard 26-letter alphabet, how many different groups of 4 letters are possible, if exactly two letters are chosen from a, e, i, o ,u and two letters are chosen from all the remaing letters except for y, and a letter can be used more than once?
A. 500 B. 1900 C. 2500 D. 3150 E. 6300
Since the order doesn’t matter and the letter can be repeated, the number of ways to choose 2 vowels is 5C2 + 5C1 = 10 + 5 = 15. (Notice that 5C2 is the number of ways to choose 2 vowels if they are different and 5C1 is the number of ways to choose 2 vowels if they are the same.)
Similarly, the number of ways to choose 2 non-y consonants is 20C2 + 20C1 = 190 + 20 = 210.
Thus, the number of ways to choose 2 vowels and 2 non-y consonants is 15 x 210 = 3150.
Answer: D
Hi, I can't get the reason for 2 vowels we use 5C2 + 5C1 = 10 + 5 = 15. I think even if a could not be repeated we would use 5C2*2! or in other words 5х4 (5 options for the first letter and 4 options for the second letter in the word).
Re: Using the standard 26-letter alphabet, how many different groups of 4
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05 Sep 2018, 01:09
Hi everyone, can someone help me to explain why it's 20C2? I thought it's supposed to be 23C2 (as y and a can be used more than once so only e, o, u are not counted).
Thank you so much
gmatclubot
Re: Using the standard 26-letter alphabet, how many different groups of 4 &nbs
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05 Sep 2018, 01:09
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26% (02:18) correct 
