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16 Feb 2016, 01:56
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John would make the 3-letter codes with 26 alphabets in condition that the middle letter must be vowel and the first letter and the third letter must be different from each other and both are consonant. How many cases of the codes are there?
A. 1,980
B. 2,020
C. 2,100
D. 2,200
E. 2,500
A. 1,980
B. 2,020
C. 2,100
D. 2,200
E. 2,500
mahakmalik
Joined: 30 Apr 2015
Last visit: 08 May 2016
Posts: 71
Own Kudos:
Given Kudos: 30
Status:Build your own dreams,Otherwise some one else will hire you to build there's.
Location: India
Concentration: Finance
Schools: BYU-Marriott'18
GMAT 1: 590 Q45 V26
GMAT 2: 660 Q47 V34
GPA: 3.68
16 Feb 2016, 20:47
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let the color code be ABC
we have 5 vowels a,e,i,o,u
B cab have any of the 5 vowels.
So,there are 5 ways
21 consonants
there are 21 ways to fill A.
as digits cannot be repeated.therefore,we can fill C in 20 ways
Totalcodes possible=21*5*20
=2100
+1 Kudo,If i helped
we have 5 vowels a,e,i,o,u
B cab have any of the 5 vowels.
So,there are 5 ways
21 consonants
there are 21 ways to fill A.
as digits cannot be repeated.therefore,we can fill C in 20 ways
Totalcodes possible=21*5*20
=2100
+1 Kudo,If i helped
17 Feb 2016, 22:56
Kudos
Bookmarks
John would make the 3-letter codes with 26 alphabets in condition that the middle letter must be vowel and the first letter and the third letter must be different from each other and both are consonant. How many cases of the codes are there?
A. 1,980
B. 2,020
C. 2,100
D. 2,200
E. 2,500
--> There should be a vowel in the middle of the 3-letter codes, which means 5 letters can be in the middle. Then, 21 letters can be in the first letter and 20 letters can be in the last letter as there should be different letters respectively. That is, 21*5*20=2,100.
Therefore, the answer is C.
A. 1,980
B. 2,020
C. 2,100
D. 2,200
E. 2,500
--> There should be a vowel in the middle of the 3-letter codes, which means 5 letters can be in the middle. Then, 21 letters can be in the first letter and 20 letters can be in the last letter as there should be different letters respectively. That is, 21*5*20=2,100.
Therefore, the answer is C.
