144 is correct.
Further Explanation:
Solution:
Number of arrangements in which no two vowels are adjacent to each other:
Given, word is ‘EAMCOT’
Vowels in word ‘EAMCOT’=
3(E, A, O) and consonants = 3(M, C, T)
Let vowels be denoted by V
Now, fix the position by Vowels like this:
The remaining 4 places can be occupied by 3 consonants
Now, arrange 3 consonants at 4 places and 3 vowels at 3 places
Formula:
Number of arrangements of n things taken all at a time = P(n, n)
∴ Total number of arrangements of vowels
= the number of arrangements of 3 things taken all at a time
= P(3, 3)
{∵ 0! = 1}
= 3!
= 3 × 2 × 1
= 6
Formula:
Number of arrangements of n things taken r at a time = P(n, r)
∴ Total number of arrangements of consonants
= the number of arrangements of 4 things taken 3 at a time
= P(4, 3)
= 4!
= 4 × 3 × 2 × 1
= 24
Proved, total number of arrangements in which no two vowels are adjacent to each other, 6 × 24 = 144
Subject: mathematics
Level: college
Keywords: solution, Vowels in word ‘EAMCOT’, formula.
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