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