NCERT Class XI Mathematics - Permutations and Combinations - Solutions
© examsnet.com
Question : 21
Total: 41
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?
(i) words start with P and end with S,
(ii) vowels are all together,
(iii) there are always 4 letters between P and S?
Solution:
There are 12 letters of which T appears 2 times
(i) When words start with P and end with S, then there are 10 letters to be arranged of which T appears 2 times.
∴ The required words =
(ii) When vowels are taken together i.e E U A I O we treat them as a single object. This single object with remaining 7 objects will account for 8 objects, in which there are 2Ts, which can be rearranged in
(iii) When there are always 4 letters between P & S
∴ P & S can be at
1st & 6th place
2nd & 7th place
3rd & 8th place
4th & 9th place
5th & 10th place
6th & 11th place
7th & 12th place.
So, P & S will be placed in 7 ways & can be arranged in 7 × 2! = 14
The remaining 10 letters with 2T’s, can be arranged in
∴ The required number of arrangements = 14 × 1814400= 25401600.
(i) When words start with P and end with S, then there are 10 letters to be arranged of which T appears 2 times.
∴ The required words =
(ii) When vowels are taken together i.e E U A I O we treat them as a single object. This single object with remaining 7 objects will account for 8 objects, in which there are 2Ts, which can be rearranged in
(iii) When there are always 4 letters between P & S
∴ P & S can be at
1st & 6th place
2nd & 7th place
3rd & 8th place
4th & 9th place
5th & 10th place
6th & 11th place
7th & 12th place.
So, P & S will be placed in 7 ways & can be arranged in 7 × 2! = 14
The remaining 10 letters with 2T’s, can be arranged in
∴ The required number of arrangements = 14 × 1814400= 25401600.
© examsnet.com
Go to Question: