I am a 10 letter word starting with I and ending with me. I have 2. I am an establishment where a hairdresser, beautician, or couturier conducts trade. I am a 5 letter word vowels and equal number of consonants. One consonant occurs twice can consecutively . the institution you study in makes you the opposite of me.
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ANSWER
'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