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11 Jan 2024, 02:30
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
46% (02:49) correct
54%
(03:10)
wrong
based on 82
sessions
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Time
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Not Attempted Yet
A 5-letter password must consist of different letters of the alphabet, with 2 vowels and 3 consonants. If the vowels must not be adjacent to each other, how many such passwords can be created?
A. 95,760
B. 159.600
C. 256.000
D. 720,000
E. 957,600
A. 95,760
B. 159.600
C. 256.000
D. 720,000
E. 957,600
Each letter of a 5-letter password must be a different letter of the alphabet. Each password must contain 2 vowels and 3 consonants, and the vowels must not be adjacent to each other. How many different passwords of this form can be made?
11 Jan 2024, 03:49
Kudos
Bookmarks
Bunuel
A 5-letter password must consist of different letters of the alphabet, with 2 vowels and 3 consonants. If the vowels must not be adjacent to each other, how many such passwords can be created?
A. 95,760
B. 159.600
C. 256.000
D. 720,000
E. 957,600
A. 95,760
B. 159.600
C. 256.000
D. 720,000
E. 957,600
Each letter of a 5-letter password must be a different letter of the alphabet. Each password must contain 2 vowels and 3 consonants, and the vowels must not be adjacent to each other. How many different passwords of this form can be made?
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Aryaa03
Joined: 12 Jun 2023
Last visit: 13 Jun 2025
Posts: 211
Given Kudos: 159
Location: India
Schools: ISB '27 (A) IIM Calcutta '26 (A) IIMA '26 (D)
GMAT Focus 1: 645 Q86 V81 DI78 
11 Jan 2024, 04:47
Kudos
Bookmarks
Bunuel
A 5-letter password must consist of different letters of the alphabet, with 2 vowels and 3 consonants. If the vowels must not be adjacent to each other, how many such passwords can be created?
A. 95,760
B. 159.600
C. 256.000
D. 720,000
E. 957,600
A. 95,760
B. 159.600
C. 256.000
D. 720,000
E. 957,600
Sol:
IMO-Option-E
Total no. of vowel : 5 , out of which 2 are to be selected : 5C2 = 10
Total no. of consonants: 21 , out of which 3 are to be selected : 21C3 = 1330
Let, x, No. of ways in which no two vowel are adjacent : Total no. of ways in which 5 selected can be arranged - Total arrangement considering with both vowel adjacent
a) Total no. of arrangement possible = 5! =120
b) Consider V1-V2 as one entity and thus, total arrangement is 4!*2 = 48 (multiplied by two as there are two ways in which V1 and V2 can be arranged)
Therefore, total no. of arrangements possible = 1330*10*(5! - 4!*2) = 13300*(120-48) = 957600
