Permutation and Combination - Quant/Math - CAT 2016Question 4 the day: September 04, 2003The question for the day is from the topic Permutation and Combination. Question: How many four letter distinct initials can be formed using the alphabets of English language such that the last of the four words is always a consonant?(1) (26^3)*(21) (2) 26*25*24*21 (3) 25*24*23*21 (4) None of these. Correct Answer - (1) Solution: The last of the four letter words should be a consonant. Therefore, there are 21 options. The first three letters can be either consonants or vowels. So, each of them have 26 options. Note that the question asks you to find out the number of distinct initials and not initials where the letters are distinct. Hence answer = 26*26*26*21 = 263 * 21 XLRI XAT TANCET Practice Question SamplesSorted by Subject | Quick LinksQuestion Archives
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CEO
Joined: 15 Aug 2003
Posts: 2955
How many four letter distinct initials can be formed using
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How many four letter distinct initials can be formed using the alphabets of English language such that the last of the four words is always a consonant?
1) (26^3)*(21)
2) 26*25*24*21
3) 25*24*23*21
4) 25^3*21
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Manager
Joined: 21 Aug 2003
Posts: 188
Location: Bangalore
[#permalink]
oops.. i am bad at grammer. what do mean by a 'consonant' ?
CEO
Joined: 15 Aug 2003
Posts: 2955
[#permalink]
Vicky wrote:
oops.. i am bad at grammer. what do mean by a 'consonant' ?
AEIOU are vowels
All other alphabets are consonants.
Intern
Joined: 11 Sep 2003
Posts: 2
Location: India
[#permalink]
3) 25*24*23*21
The last letter can be choosen in 21 ways ( 26 consonants - 5 vowels).
Having choosen the last letter, we can chhose the other three letters among 25 letters in 25*24*23
Hence the total number of ways we canchoose is 25*24*23*21
Correct me , if I am wrong.
Intern
Joined: 16 Jul 2003
Posts: 16
[#permalink]
A.
There are no restrictions about repetition. So total possible words with 3 alphabets = 3^26. multiply this by 21( last letter
can only be consonant).
Pls. confirm the answer.
CEO
Joined: 15 Aug 2003
Posts: 2955
[#permalink]
random_choice wrote:
3) 25*24*23*21
The last letter can be choosen in 21 ways ( 26 consonants - 5 vowels).
Having choosen the last letter, we can chhose the other three letters among 25 letters in 25*24*23
Hence the total number of ways we canchoose is 25*24*23*21
Correct me , if I am wrong.
The question asks you to find out the number of distinct initials and not initials where the letters are distinct.
First letter can be arranged in 26 ways
Second letter can be arranged in 26 ways
Third letter can be arrranged in 26 ways
last one...NO vowels...so 26- 5 =21 ways
26^3 * 21 ways
Answer A
thanks all
praetorian
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