In how many ways can the letters of the word PERMUTATIONS be arranged if the
(i) Words start with P and end with S
(ii) Vowels are all together
(iii) There are always 4 letters between P and S
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(i) Total number of letters in PERMUTATIONS =12
The only repeated letter is T; 2times
The first and last letters of the word are fixed as P and S, respectively.
Number of letters remaining =12 – 2 = 10
⇒ Number of permutations =
(ii) Number of vowels in PERMUTATIONS = 5 (E, U, A, I, O)
Now, we consider all the vowels together as one.
Number of permutations of vowels = 120
Now total number of letters = 12 – 5 + 1= 8
⇒ Number of permutations =
Therefore, total number of permutations = 120 × 20160 = 2419200
(iii) Number of places is as 1 2 3 4 5 6 7 8 9 10 11 12
There should always be 4 letters between P and S.
Possible places of P and S are 1 and 6, 2 and 7, 3 and 8, 4 and 9, 5 and 10, 6 and 11, 7 and 12
Possible ways =7
Also, P and S can be interchanged.
No. of permutations =2 × 7 =14
The remaining 10 places can be filled with 10 remaining letters,
∴ No. of permutations =
Therefore, the total number of permutations = 14 × 1814400 =25401600
Access other exercise solutions of Class 11 Maths Chapter 7 – Permutations and CombinationsExercise 7.1 Solutions 6 Questions
Exercise 7.2 Solutions 5 Questions
Exercise 7.4 Solutions 9 Questions
Miscellaneous Exercise on Chapter 7 Solutions 11 Questions
Also explore – NCERT Class 11 Solutions