Question

Find the number of 7 lettered words each consisting of 3 vowels and 4 consonants which can be formed using the letters of the word “DIFFERENTIATION”.

Answer 1

Hint: 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 of selection where the order of selection is a factor.
In this question, we need to determine the total number of 7 lettered words that can be formed using the word “DIFFERENTIATION so that ways so that the word consists of 3 vowels and 4 constants. For this, we need to follow the concept of combinations by separating the vowels and constant and placing them one by one.


Complete step by step solution:
The given word is “DIFFERENTIATION.”
In the word, there are 7 vowels letters ‘IIIEEAO’ and 8 contents letters ‘DFFRNTTN’
In vowels, the number of letters
‘I’ is three times
‘E’ is two times
‘A’ is one
‘O’ is one
Now we have to choose 3 vowels,
Case 1: When all vowels are the same, this is possible for ‘I’ only, then the number of combinations is
1--(i)
Case 2: When 2 vowels are the same and 1 different, this is possible for two ‘I’s, or two ‘E’s and one ‘A’, ’O’ then, the number of combination is
 $
  {}^3{C_1} \times {}^3{C_2} \times {}^2{C_1} = \dfrac{{3!}}{{(3 - 1)!1!}} \times \dfrac{{3!}}{{(3 - 2)!2!}} \times \dfrac{{2!}}{{(2 - 1)!1!}} \\
   = 3 \times 3 \times 2 \\
   = 18 - - - - (ii) \\
 $
Case 3: All the three vowels are unique (or different) then, the number of combinations is:
$
  3! \times {}^4{C_3} = 6 \times \dfrac{{4!}}{{(4 - 3)!3!}} \\
   = 6 \times 4 \\
   = 24 - - - - (iii) \\
 $
Hence, the total number of combinations for arranging three vowels from the word “DIFFERENTIATION” is $1 + 18 + 24 = 43 - - - - (iv)$
Similarly, for the arrangement of four consonants out of 8 contents letters ‘DFFRNTTN’ in which
‘D’ is 1 time
‘F’ is 2 times
‘R’ is 1 time
‘N’ is 2 times
‘T’ is 2 times
Case 1: Two times same consonants then, the combination is given as:
$
  {}^3{C_2} \times {}^4{C_2} = \dfrac{{3!}}{{(3 - 2)!2!}} \times \dfrac{{4!}}{{(4 - 2)!2!}} \\
   = 3 \times 6 \\
   = 18 - - - - (v) \\
 $
Case 2: Two same consonants and two unique consonants the, the combination is given as:
$
  2! \times {}^3{C_1} \times {}^4{C_2} \times {}^4{C_2} = 2 \times \dfrac{{3!}}{{(3 - 1)!1!}} \times \dfrac{{4!}}{{(4 - 2)!2!}} \times \dfrac{{4!}}{{(4 - 2)!2!}} \\
   = 2 \times 3 \times 6 \times 6 \\
   = 216 - - - - (vi) \\
 $
Case 3: All the four consonants are unique (or different) then, the number of combination is given as:
$
  4! \times {}^5{C_4} = 4 \times 3 \times 2 \times \dfrac{{5!}}{{(5 - 4)!4!}} \\
   = 24 \times 5 \\
   = 120 - - - - (vii) \\
 $
Hence, the total number of combinations for arranging three consonants from the word “DIFFERENTIATION” is $18 + 216 + 120 = 354 - - - - (viii)$
Now, selecting three positions to place the vowels, the total number of words thus equals to $
  43 \times 354 \times {}^7{C_3} = 43 \times 354 \times \dfrac{{7!}}{{(7 - 3)!3!}} \\
   = 15222 \times \dfrac{{7 \times 6 \times 5 \times \left( {4!} \right)}}{{4!3!}} \\
   = 15222 \times 35 \\
   = 532770 \\
 $
Hence, the number of 7 lettered words each consisting of 3 vowels and 4 consonants which can be formed using the letters of the word “DIFFERENTIATION” is 532770.


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