Question

How many strings are there using the letters of the word INTERMEDIATE, if the vowels and consonants are alternative

Answer 1

The given word is INTERMEDIATE

Number of letters = 12

Number of I’S = 2

Number of T’S = 2

Number of E’S = 3

Vowels are A, I, I, E, E, E

Total number of vowels = 6

Consonants are N, T, R, M, D, T

Total number of consonants = 6

Vowels and consonants are alternative

V	C	V	C	V	C	V	C	V	C	V	C
1	2	3	4	5	6	7	8	9	10	11	12

(a) Let the first box be filled with a vowel.

There are six alternate places available for 6 vowels.

∴ Number of ways of filling 6 vowels in the alternative six boxes is `(6!)/(2! xx 3!)`

Remaining 6 boxes can be filled with the 6 consonants.

Number of ways of filling the 6 consonants in the remaining 6 boxes is `(6!)/(2!)`

Total number of ways = `(6!)/(2! xx 3!) xx (6!)/(2!)`

C	V	C	V	C	V	C	V	C	V	C	V
1	2	3	4	5	6	7	8	9	10	11	12

(b) Let the first box be filled with a consonant.

There six alternate places available for 6 consonants.

∴ Number of ways of filling 6 consonants in the six alternate boxes is `(6!)/(2!)`

Remaining 6 boxes can be filled with the 6 vowels

∴ Number of ways of filling the 6 vowels in the remaining 6 boxes is `(6!)/(2! xx 3!)`

Total number of ways = `(6!)/(2!) xx (6!)/(2! xx 3!)`

∴ Total number of strings formed by using the letters of the word INTERMEDIATE, if the vowels and consants are alternative

= `(6!)/(2! xx 3!) xx (6!)/(2!) + (6!)/(2!) xx (6!)/(2! xx 3!)`

= `2 xx (6! xx 6!)/(2! xx3! xx 2!)`

= `(2 xx 6 xx 5 xx 4 xx 3 xx 2 xx 1 xx 6 xx 5 xx 4 xx 3 xx 2 xx 1)/(1 xx 2 xx 1 xx 2 xx 3 xx 1 xx 2)`

= 60 × 720

= 43200