Question

I am a five letter word with 2 vowels , both vowels same . I start with R , and end with W , i am known for starting something again or bringing back to an original condition .l​

Answer 1

Answer:

Explanation:

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 H 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+!=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!-25929

=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×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

Th the same way, 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

3

2

C ways. Selecting 5 consonants can be done in

5

3

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 ×

2

3

C

= 3 × 10 × 20

= 3600 ways

Answer 2

Answer:

Renew

Explanation:

hope it helps.............