The number of arrangements that can be formed out of “LOGARITHM” so that
[1] No two vowels come together is
[A] $6!\,7{p_3}$
[B] $6!\,7!$
[C] $6!\,3!$
[D] $7!\,3!$
[2] All the vowels not come together is
[A] $6!\,7{p_3}$
[B] $9! - 7!\,3!$
[C] $7!\,3!$
[D] $6!\,7!$
[3] No two consonants come together is
[A] $0$
[B] $6!\,7!$
[C] $6!\,3!$
[D] $7!\,3!$
Answer
Hint: Here, the given question is from permutation and combination. We have to find the number of arrangements that can be formed out by the letters of the word “LOGARITHM” considering different conditions. So, firstly we have to count the number of consonants and vowels in the word “LOGARITHM”, then apply the conditions such as in the first question arrange the consonant by putting a space between two consonants and then arrange the vowels in that place.
Complete step-by-step
solution:
[1] Here, the letters of the word “LOGARITHM” are arranged in such a way that no two vowels come together.
Since, the given condition is that no two vowels come together and total six consonants and three vowels are in the word “LOGARITHM” so the possible arrangement may be VCVCVCVCVCVCV. And of the six positions of consonant, six consonants can be arranged in $6!$ ways. The arrangement of six consonants produces seven places for vowels and we have only three vowels so this
can be arranged as $7{p_3}$.
So, the total possible number of arrangements is $6!\,7{p_3}$.
Hence, option [A] is correct.
[2] Here, the letters of the word “LOGARITHM” are arranged in such a way that all the vowels do not come together.
To arrange according to given condition, find the total possible number of arrangements of letters without any condition that is $9!$ and then subtract the case when all the vowels come together from this.
Now, put all vowels in to a box and
consider it as a single letter and make arrangements of six consonants and a box of vowels that comes as $7!\,3!$ because the seven letters can be arranged in $7!$ and the vowels inside the box can be arranged internally in $3!$ ways.
So, the total possible number of arrangements is $9! - 7!\,3!$
Hence, option [B] is correct.
[3] Here, the letters of the word “LOGARITHM” are arranged in such a way that no two consonants come together.
We have only three vowels but six
consonants and we have to arrange such that no two consonants come together and this can be possible if we arrange the letters as CVCVCVC but there is only four places for consonants but we have to arrange six consonants so, it is not possible to arrange in such a way.
So, the possible number of arrangements is zero.
Hence, option [A] is correct.
Note: We should know,
$n{p_r} = \dfrac{{n!}}{{\left[ {n - r} \right]!\,}}$
\[n{C_r} = \dfrac{{n!}}{{\left[ {n - r}
\right]!\,r!}}\]
It will be helpful in finding the numerical value.
- Forum
- Algebra
- Permutations
Chilukuri Sai Kartik,
12 years ago
Grade:12
FOLLOW QUESTION
We will notify on your mail & mobile when someone answers this question.
Enter email id Enter mobile number
1 Answers
abhishek athreya
19 Points
12 years ago
in the word LOGARITHM
the position of vowels is 2,4,6
the position of consonants is 1,3,5,7,8,9
as their relative positions should be maintained
the vowels can occupy only positions 2,4,6
and the consonants can only occupy positions 1,3,5,7,8,9
the permutation among vowels = 3!
the permutation among consonants =6!
thus total permutation = 6!*3!
=4320
Think You Can Provide A Better Answer ?
Provide a better Answer & Earn Cool Goodies See our forum point policy
View courses by askIITians
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !
Select Grade Select Subject for trial