Question

Number of words each consisting of two vowels and two consonants which can be made out of the letters of the word ‘DEVASTATION’ is

1. 126
2. 198
3. 1512
4. 1638

Answer 1

Use the method of permutation and combination. A permutation is the number of ways in which objects from a set may be selected, generally without replacement to form a subset, whereas combination is the method in which objects are selected depending on the factor of order of selection.
In this question, we need to determine the number of words each consisting of two vowels and two consonants which can be made out of the letters of the word ‘DEVASTATION’ so that the word consists of 2 vowels and 2 constants. For solving this, we need to use the concept of combinations by separating the vowels and constants and then placing them one by one.

Complete step by step solution:
The given word is ‘DEVASTATION’
In the word, there are 5 vowels letters ‘EAAIO’ and 6 contents letters ‘DVSTTN’
In vowels, the number of letters ‘I’ is one time, ‘E’ is one time, ‘A’ is two times and ‘O’ is one time. Now we have to choose 2 vowels,
Case 1: When all vowels are the same, this is possible for ‘A’ only, then the number of combinations is 1-(i)
Case 2: When 2 vowels are the different (or unique), then, the number of combinations is:$2!\,{ \times ^4}{C_2} = 2 \times \dfrac{{4!}}{{(4 - 2)!2!}} = 12$ - - - - (ii)
Hence, the total number of combinations for arranging two vowels from the word ‘DEVASTATION’ is 1 + 12 = 13 - - - - (iii)
Similarly, for the arrangement of four consonants out of 6 contents letters ‘DVSTTN’ in which
‘D’ is 1 time
‘V’ is 1 time
‘S’ is 1 time
‘T’ is 2 times
‘N’ is 1 time
Case 1: Two same consonants, the combination is given as: 1 ------ (iv)
Case 2: Two consonants are unique (or different) then, the number of combinations is given as:
$2!{ \times ^5}{C_2} = 2! \times \dfrac{{5!}}{{(5 - 2)!2!}} = 5 \times 4 = 20$- - - - (v)
Hence, the total number of combinations for arranging three consonants from the word ‘DEVASTATION’ is 1 + 20 = 21 - - - - (vi)
Now, selecting two positions to place the vowels, the total number of words thus equals to
$= 13 \times 21 \times {}^4{C_2} = 13 \times 21 \times \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}} = 1638$
Hence, the number of words each consisting of two vowels and two consonants which can be made out of the letters of the word ‘DEVASTATION’ is 1638.

Therefore, option (D) is the correct answer.

Note:
In this question, students must take care that here the vowels as well as the consonants are repeating in nature as the word ‘DEVASTATION’ has more than one similar vowel and consonants and so we need to apply the arrangement in that also.