How many different ways can the letters of the word logarithm be arranged such that vowels are never together?

Exercise :: Permutation and Combination - General Questions

13.

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

A.	10080
B.	4989600
C.	120960
D.	None of these

Answer: Option C

Explanation:

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

Number of ways of arranging these letters =	8!	= 10080.
(2!)(2!)

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters =	4!	= 12.
2!

Required number of words = (10080 x 12) = 120960.

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Exercise :: Permutation and Combination - General Questions

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

Answer: Option D

Explanation:

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

Required number of numbers = (1 x 5 x 4) = 20.

Answer

Verified

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.

The letters of the word LOGARITHM are arranged at random. Find the probability that vowels are never together.

There are 9 letters in the word LOGARITHM.
These letters can be arranged among themselves in 9P9 = 9! ways∴ n(S) = 9!There are 9 letters in the word LOGARITHM. These letters can be arranged among themselves in 9 P9 = 9! ways

C C C C C C

6 consonants create 7 gaps.

∴ 3 vowels can be arranged in 7 gaps in 7P3 ways. Also 6 consonants can be arranged among themselves in 6P6 = 6! ways.

∴ n(B) = 6! x `""^7"P"_3`

∴ P(B) = `("n"("B"))/("n"("S")`

= `(6! xx""^7"P"_3)/(9!)`

Concept: Elementary Properties of Probability

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Page 2

There are 9 letters in the word LOGARITHM.
These letters can be arranged among themselves in 9P9 = 9! ways.∴ n(S) = 9!Let C be the event that exactly 4 letters are arranged between G and H.Consider the following arrangement1 2 3 4 5 6 7 8 9∴ Out of 9 places, G and H can occupy any one of the following 4 positions in 4 ways.

1st and 6th, 2nd and 7th, 3rd and 8th, 4th and 9th

Now, G and H can be arranged among themselves in 2P2 = 2! = 2 ways.
Also, the remaining 7 letters can be arranged in the remaining 7 places in 7P7 = 7! ways.
∴ n(C) = 4 × 2 × 7! = 8 × 7! = 8!

∴ P(C) = `("n"("C"))/("n"("S")`

= `(8!)/(9!)`

= `(8!)/(9xx8!)`

= `1/9`

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The letters of the word LOGARITHM are arranged at random. Find the probability that Begins with O and ends with T.

There are 9 letters in the word LOGARITHM.
These letters can be arranged among themselves in 9P9 = 9! ways.∴ n(S) = 9!Let D be the event that word begins with O and ends with T.

Thus first and the last letter can be arranged in one way each and the remaining 7 letters can be arranged in the remaining 7 places in 7P7 = 7! ways

∴ n(D) = 1 × 7! × 1 = 7!

∴ P(D) = `("n"("D"))/("n"("S")`

= `(7!)/(9!)`

Concept: Elementary Properties of Probability

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Page 4

The letters of the word LOGARITHM are arranged at random. Find the probability that start with vowel and end with ends with consonant.

There are 9 letters in the word LOGARITHM.
These letters can be arranged among themselves in 9P9 = 9! ways.∴ n(S) = 9!Let E be the event that word starts with vowel and ends with consonant.There are 3 vowels and 6 consonants in the word LOGARITHM.∴ The first place can be filled in 3 different ways and the last place can be filled in 6 ways.

Now, remaining 7 letters can be arranged in 7 places in 7P7 = 7! ways

∴ n(E) = 3 × 7! × 6

∴ P(E) = `("n"("E"))/("n"("S")`

= `(3 xx 7! xx 6)/(9!)`

Concept: Elementary Properties of Probability

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Page 5

The word SAVITA contains 6 letters. Out of 6 letters, 3 are vowels (A, A, I) and 3 are consonants (S, V, T).
6 letters in which A repeats twice can be arranged among themselves in `(6!)/(2!)` ways.

∴ n(S) = `(6!)/(2!)`Let A be the event that vowels are always together.3 vowels (A, A, I) can be arranged among themselves in `(3!)/(2!)` ways.

Considering 3 vowels as one group, 3 consonants and this group (i.e. altogether 4) can be arranged in 4P4 = 4! ways.

∴ n(A) = `4! xx (3!)/(2!)`

∴ P(A) = `("n"("A"))/("n"("S"))`

= `(4!xx(3!)/(2!))/((6!)/(2!))`

= `(4!3!)/(6!)`

= `1/5`