In how many different ways can the letters in the word "HYDERABAD" be arranged so that all the vowels come together and all the consonants come together?
Answers
Answer:
Permutations and Combinations: In how many ways can the letters of the word 'interference' so that no two consonant are together?
Explanation:
Permutations and Combinations: In how many ways can the letters of the word 'interference' so that no two consonant are together?
The word 'EQUATION' has 8 letters. It has 5 vowels (A E I O U) AND 3 consonants (Q T N) in it and these 5 vowels should
always come together AND 3 consonants come together.
Hence these 5 vowels can be grouped and considered as a single
letter. That is, (A E I O U).
Hence these 3 consonants can be grouped and considered as a single
letter. That is, (Q T N).
Total = (A E I O U) (Q T N)
Hence we can assume total letters as 2 and all these letters are different.
Number of ways to arrange these letters = 2!= 2 x 1 = 2
In the 5 vowels (A E I O U), all the vowels are different.
Number of ways to arrange these vowels among themselves = 5! = 5x4x3x2 x 1 = 120
In the 3 consonants (Q T N), all the consonants are different.
Number of ways to arrange these consonants among themselves = 3! =x3x2 x 1 = 6
Total number of ways = 2x120x6 = 1440
In the word 'equation', there are five vowels (a,e, i, o, u) and three consonants (q, t, n).
Since the vowels and consonants should always occur together, we have to group them and consider as two separate things, say V and C.
Now V and C can be arranged in two ways as VC or CV. In both these cases, the vowels and consonants are not separated.
Since the vowels are all distinct, they can be arranged in 5! ways.
Similarly, as the consonants are also distinct, they can be arranged in 3! ways.
Therefore, the total number of arrangements is 2*5!*3!
2*120*6 = 1440
1440 ways
Split the word equation into two mutually exclusive and exhaustive sets :- Vowels : {e,u,a,i,o} and Consonants : {q,t,n} (Since you want vowels and consonants always occur together) . Now either vowels' set occur first and then consonants' or vice-verse. So for this you have 2! ways = 2 ways. Now elements within vowels' set can be permuted in 5! ways = 120 ways and elements within consonants' set can be permuted in 3! ways = 6 ways. So completing your job requires and (conjunction) of these 3 sub-jobs. Hence total ways = 2!*5!*3! = 1440.