Question

Find the number of words those can be formed by using all letters of the word ' DAUGHTER'. If
(i) Vowels occurs in first and last place.
(ii) Start with letter $G$ and end with letters $H$.
(iii) Letters $G,H,T$ always occurs together.
(iv) No two letters of $G,H,T$ are consecutive
(v) No vowel occurs together
(vi) Vowels always occupy even place.
(vii) Order of vowels remains same.
(viii) Relative order of vowels and consonants remains same.
(ix) Number of words are possible by selecting $2$ vowels and $3$ consonants.

Answer 1

'DAUGHTER' :Total words=8

Vowels are 'AUE'

Consonants are 'DGHTR'

(1) Vowels occur in first and last place

For the first letter, there are 3 possibilities

so, after filing the first letter, the last letter has 2 possibilities

Filing the other 6 places, we can do that in 6! ways because all letters are different.

So, the number of ways = 3*2*6!

=4320 ways

(2)Start with letter G and end with letter H

Remaining places can be filled in 6! ways=720 ways.

(3)Letters G, H, T occurs together always

Consider GHT as one object

then the number of object = 5+1=6

Numbers of arranging them=6! ways

We can re-arrange G, H, T among themselves in 3! ways.

So, the number of ways = 3!6!=4320 ways.

(4)No 2 letters G, H, T are consecutive

case(1): All 3 are consecutive

Number of ways=4320 ways

case(2): Only G, H are consecutive

Consider GH as one object

then the number of objects = 6+1=7

Numbers of arranging them=7!2! ways

But in these, we also get the cases in which G, H, T come together

so we need to subtract 4*6!

The number of arrangements in which the only G, H are consecutive= 7!2!-4*6!

Total number of ways in which at least 2 of G, H, T are consecutive= 4320+3(7!2!-4*6!)=25920

Number of ways of arranging such that no two ways in which at least 2 of G, H, T are consecutive

=Number of ways-25920

=8!-25920

=14,400 ways

(5) No vowel occur together

3 vowels- A, U, E.

The same argument as above follows for this.

So, number of ways= 14,400

(6) Vowels occupy even space

There are 4 even places and 3 vowels

'so we can arrange vowels in ${}_3^{4}P$ ways.

The remaining 5 letters can be arranged in 5! ways.

So, number of ways= ${}_3^{4}P\times 5!$ =2880 ways,

(7) order of vowels remains same

The total number of arrangements =8! ways

But in these, we cannot change the order of vowels.

Suppose the required ways be 'n' multiplied by 3! ways

with 'n' must be given 8!

i.e 3!.n=8!

n=6720 ways.

(8) order of vowels and consonants remains same

Here, there are 5 consonants and 3 vowels

i.e 5!3!n=8!

n=56 ways.

(9) We need to select two vowels which can be done in ${}_2^{3}C$ ways. Selecting 5 consonants can be done in ${}_3^{5}C$ ways.

Rearranging can be done 5! ways

So,number of words that are possible by selecting 2 vowels and 3 consonants

$={}_3^{5}C\times {}_2^{3}C\times 5!$

$=3600$ ways