Question

Determine how many words can be formed using the letters of the word LOGARITHM. if
(a) vowels are always together.
(b) no two vowels are together.
(c) consonants occupy even positions.
(d) begin with O and end with T.

Answer 1

Answer:

(a) 30240

(b) 151200

(c) 0

(d) 5040

Step-by-step explanation:

Hi,

Given word LOGARITHM, which consists of 9 letters

of which all 9 are distinct.

There are 3 vowels and 6 consonants in the word.

(a) Vowels are always together

If vowels are always together , we can treat all 3 vowels

as 1, and the rest 6 distinct letters, in total all these 7

can be arranged in 7! ways. But all 3 vowels which are

together as 1  thy can permute among themseleves

in 3! ways. Hence, total number of ways in which vowels

are always together are 7!*3!

= 5040*6

= 30240

(b) no two vowels are together

Firstly place all the 6 consonants,

All 6 consonants can be arranged in 6! ways.

Now, let us place 3 vowels in between these consonants,

so that no 2 vowels will be together.

After placing 6 consonants, there are 5 positions in between

the 6 consonant and 2 end points on either side of starting

and ending consonant, so all vowels can be placed in these

7 places. Since there are 3 vowels, we need to chose 3 places

to fix these vowels, which can be done in ⁷C₃ ways = 35

And all these 3 vowels can interchange among themselves

in 3! ways = 6. Hence total number of ways in which no 2

vowels are together are 6!*35*6

= 151200

(c) consonants occupy even position

There are 9 letters in total , so there are only 4 even positions

But the number of consonants are 6 in number, So, this type

of arrangement is not possible.

Hence , total number of ways in which consonants occupy even

positions are 0.

(d) begin with O and end with T

Given that first letter of the word is O and the last letter is T,

so rest of the 7 letters can be arranged in between in 7! ways

= 5040

Hope, it helps !