Question

The number of six letter words, each consisting of three vowels and three consonants, that can be formed using the letters of the word ‘CIRCUMFERENCE’ is
A) 22100
B) 22150
C) 22101
D) 21200

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 in which objects are selected depending on the factor of order of selection.
In this question, we need to determine the number of six letter words each consisting of three vowels and three consonants which can be made out of the letters of the word ‘CIRCUMFERENCE’. 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 ‘CIRCUMFERENCE’
In the word, there are 5 vowels letters ‘IUEEE’ and 8 contents letters ‘CRCMFRNC’
In vowels, the number of letters ‘I’ is one time, ‘E’ is three times and ‘U’ is one time. Now we have to choose 3 vowels,
Case 1: When all vowels are the same, this is possible for ‘E’ only, then the number of combinations is 1-(i)
Case 2: When 2 vowels are the same and third is different, this is possible for two ’E’s and one ‘I’ or one ‘U’, then, the number of combinations is:\[^3{C_1}{ \times ^2}{C_1} = \dfrac{{3!}}{{(3 - 1)!1!}} \times \dfrac{{2!}}{{(2 - 1)!1!}} = 6\] - - - - (ii)
Case 3: When all the three vowels are different, then, the number of combinations is:\[2!\,{ \times ^3}{C_2} = 2 \times \dfrac{{3!}}{{(3 - 2)!2!}} = 6\] - - - - (iii)
Hence, the total number of combinations for arranging three vowels from the word ‘CIRCUMFERENCE’ is 1 + 6 + 6 = 13 - - - - (iv)
Similarly, for the arrangement of four consonants out of 8 contents letters ‘CRCMFRNC’ in which
‘C’ is 3 times
‘R’ is 2 times
‘M’ is 1 time
‘F’ is 1 time
‘N’ is 1 time
Case 1: All three same consonants, the combination is given as: 1 ------ (v)
Case 2: Two consonants are same and one different then, the number of combinations is given as:
$^2{C_1}{ \times ^4}{C_1} \times 3 = \dfrac{{2!}}{{(2 - 1)!1!}} \times \dfrac{{4!}}{{(4 - 1)!1!}} \times 3 = 2 \times 4 \times 3 = 24$- - - - (vi)
Case 2: All consonants are different then, the number of combinations is given as:
$^5{C_3} \times 6 = \dfrac{{5!}}{{(5 - 3)!3!}} \times 6 = 10 \times 6 = 60$- - - - (vi)
Hence, the total number of combinations for arranging three consonants from the word ‘CIRCUMFERENCE’ is 1 + 24 + 60 = 85 - - - - (vi)
Now, selecting three positions to place the vowels, the total number of words thus equals to
$ = 13 \times 85 \times {}^6{C_3} = 13 \times 21 \times \dfrac{{6!}}{{3!\left( {6 - 3} \right)!}} = 22100$
Hence, the number of six letter words, each consisting of three vowels and three consonants, that can be formed using the letters of the word ‘CIRCUMFERENCE’ is 22100.

Therefore, option (A) 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 ‘CIRCUMFERENCE’ has more than one similar vowel and consonants and so we need to apply the arrangement in that also.