Question 1 :
How many strings are there using the letters of the word INTERMEDIATE, if
(i) The vowels and consonants are alternative
(ii) All the vowels are together
(iii) Vowels are never together
(iv) No two vowels are together.
Solution :
Number of letters in the given word "INTERMEDIATE"
= { I, N, T, E, R, M, E, D, I, A, T, E } = 12
Here "E" is repeating 3 times, "I" is repeating 2 times, "T" is repeating 2 times.
Vowels = {A, E, E, E, I, I}
Number of vowels = 6
Consonants = {N, T, T, R, D, M}
Number of consonants = 6
Case (i)
If vowels will occupy odd places, then consonants will occupy even places.
Case (ii)
If vowels will occupy even places, then consonants will occupy odd places.
Number of ways to fill odd places = 6!/3!2!
= (6 ⋅ 5 ⋅ 4)/(2 ⋅ 1)
= 3 ⋅ 5 ⋅ 4 = 60
Number of ways to fill even places = 6!/2!
= 6 ⋅ 5 ⋅ 4 ⋅ 3
= 360
Number of ways in case (i) = 60 ⋅ 360 = 21600
Number of ways in case (ii) = 60 ⋅ 360 = 21600
Total ways = 21600 + 21600 = 43200
Since all vowels are together, we have to take it as one unit.
Considering the consonants as independent things. So, we have 7 units. Vowels may shuffle them among themselves in 6 places.
So, number of ways = (7! ⋅ 6!) / (3!⋅2!⋅2!)
= (7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1) ⋅ (6 ⋅ 5)
= 151200
Hence the total ways is 151200.
To find the answer for the above case, let us follow the step given below.
Vowels are never together
= (Number of ways of arranging the letters of the given word without restriction) - (Number of ways of arranging the the letters with vowels together)
= [12!/ (3!⋅2!⋅2!)] - 151200
= 19958400 - 151200
= 19807200
No two vowels are together means, we may fix vowels and consonants in alternate places and any two consonants may be placed between two vowels.
Note : We cannot take more than two consonants at a time.
= 21600 + 21600 + 108000
= 151200
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