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08 May 2020, 03:43
Kudos
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C
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:
43% (02:04) correct
57%
(02:54)
wrong
based on 100
sessions
History
Date
Time
Result
Not Attempted Yet
If all words with 2 distinct vowels and 3 distinct consonants were listed alphabetically, what would be the rank of “ACDEF’?
A. 1717
B. 4716
C. 4718
D. 4720
E. 5624
Are You Up For the Challenge: 700 Level Questions
A. 1717
B. 4716
C. 4718
D. 4720
E. 5624
Are You Up For the Challenge: 700 Level Questions
08 May 2020, 04:26
Kudos
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The words that comes ahead of “ACDEF’ start with AB, ACB, ACDB and ACDEB
The number of words that start with AB = 4C1 * 20C2 *3! = 4560
(We have to select 1 vowel out of remaining 4 and 2 consonants out of remaining 20. The number of ways to arrange last 3 alphabets is 3!, as the first 2 are fixed)
The number of words that start with ACB = 4C1 * 19C1 *2! = 152
(We have to select 1 vowel out of remaining 4 and 1 consonant out of remaining 19. The number of ways to arrange last 2 alphabets is 2!, as the first 3 are fixed)
The number of words that start with ACDB = 4C1*1! = 4
(We have to select 1 vowel out of remaining 4 )
Total word that comes ahead of “ACDEF" = 4560+152+4+1= 4717
Rank= 4717+1= 4718
The number of words that start with AB = 4C1 * 20C2 *3! = 4560
(We have to select 1 vowel out of remaining 4 and 2 consonants out of remaining 20. The number of ways to arrange last 3 alphabets is 3!, as the first 2 are fixed)
The number of words that start with ACB = 4C1 * 19C1 *2! = 152
(We have to select 1 vowel out of remaining 4 and 1 consonant out of remaining 19. The number of ways to arrange last 2 alphabets is 2!, as the first 3 are fixed)
The number of words that start with ACDB = 4C1*1! = 4
(We have to select 1 vowel out of remaining 4 )
Total word that comes ahead of “ACDEF" = 4560+152+4+1= 4717
Rank= 4717+1= 4718
Bunuel
08 May 2020, 07:09
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Bunuel
Solution
- • If listed alphabetically, all the words whose starting letters are “AB”, “ACB”, “ACDB”, “ACDEB” will come before the letter “ACDEF”
• Since, there are 5 vowels and 21 consonants.
- o So, total number of words that starts with “AB” and has 2 distinct vowel and 3 distinct consonants \(= (5-1)C1* (21-1)C2*3! = 4C1*20C2*3! = 4*10*19*6=4560\)
o total number of words that starts with “ACB” and has 2 distinct vowel and 3 distinct consonants \(= (5-1)C1* (21-2)C1*2! = 4C1*19C1*2! = 4*19*2 = 152\)
o total number of words that starts with “ACDB” and has 2 distinct vowel and 3 distinct consonants \(= (5-1) C1 =4C1 = 4\)
• So rank of the word “ACDEF” \(= (4560 + 152 + 4+ 1) + 1 = 4718\)
