Chapter 7 Permutations and Combinations Ex 7.3 Solutions
Question - 11 : - 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?
Answer - 11 : -
(i) Total number of letters in PERMUTATIONS =12
Only repeated letter is T; 2times
First and last letter of the word are fixed as P and S respectively.
Number of letters remaining =12 тАУ 2 = 10
тЗТ Number of permutations =
┬а
(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 тАУ 5 + 1= 8
тЗТ Number of permutations =
┬а
Therefore, total number of permutations = 120 ├Ч 20160 = 2419200
(iii) Number of places are 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, 2and 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 ├Ч 7 =14
Remaining 10 places can be filled with 10 remaining letters,
тИ┤ No. of permutations =
┬а
Therefore, total number of permutations = 14 ├Ч 1814400 =25401600.