Showing posts with label Words. Show all posts
Showing posts with label Words. Show all posts
Monday, 13 May 2019
Sunday, 12 May 2019
Aptitude Hack#79(13-5-19)
Question:
How many words can be formed by using all letters of the word “Bihar”
A ) 110
B ) 120
C ) 130
D ) 140
Answer: B) 120
Explanation:
The word Bihar contains 5 different letters.
Required number of words = 5p5
= 5! = (5x4x3x2x1)
= 120.
How many words can be formed by using all letters of the word “Bihar”
A ) 110
B ) 120
C ) 130
D ) 140
Answer: B) 120
Explanation:
The word Bihar contains 5 different letters.
Required number of words = 5p5
= 5! = (5x4x3x2x1)
= 120.
Sunday, 5 May 2019
Tuesday, 30 April 2019
Aptitude Hack#72(6-5-19)
Question:
How many words can be formed by using all letters of the word ‘daughter’ so that the vowels always come together?
A ) 4320 words
B ) 4220 words
C ) 4120 words
D ) 5320 words
Answer: A) 4320 words
Explanation:
Given the word contains 8 different letters.
When the vowels aue are always together,
we may suppose them to form an entity, treated as one letter.
Then, the letters to be arranged are dgntr (aue).
Then 6 letters to be arranged in 6p6 = 6! = 720 ways.
The vowels in the group (aue) may be arranged in 3! = 6 ways.
Required number of words = (720×6) = 4320.
How many words can be formed by using all letters of the word ‘daughter’ so that the vowels always come together?
A ) 4320 words
B ) 4220 words
C ) 4120 words
D ) 5320 words
Answer: A) 4320 words
Explanation:
Given the word contains 8 different letters.
When the vowels aue are always together,
we may suppose them to form an entity, treated as one letter.
Then, the letters to be arranged are dgntr (aue).
Then 6 letters to be arranged in 6p6 = 6! = 720 ways.
The vowels in the group (aue) may be arranged in 3! = 6 ways.
Required number of words = (720×6) = 4320.








