Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
6.6k views
in Permutations by (46.2k points)
closed by

Suppose the word ‘PENCIL’ is given to us and we have to form words with the letters of this word.

1 Answer

+1 vote
by (49.2k points)
selected by
 
Best answer

1. There is no restriction on the arrangement of the letters. 

The six different letters can be arranged in 6P6 = 6 ! = 720 ways. 

Hence the total number of words formed = 720. 

2. All words begin with a particular letter. 

Suppose all words begin with E. The remaining 5 places can be filled with remaining 5 letters in 5 ! = 120 ways. 

3. All words begin and end with particular letters. 

Suppose all words begin with L, and end with P. The remaining 4 places can then be filled in 4 ! ways. ∴ The total number of words formed = 4 ! = 24.

Note. If the words were to begin or end with E or L, these two positions could have been filled in 2P2 = 2 ways. Hence the number of words in this case would have been = 2 × 24 = 48.

4. N is always next to E.

P EN C I L

Since N is always next to E, therefore ‘EN’ is considered to be one letter. 

∴ Required no. of permutations = 5! = 5 × 4 × 3 × 2 × 1 = 120. 

5. Vowels occur together: The vowels are E and I. Regarding them as one letter, the 5 letters can be arranged in 5 ! = 120 ways. These two vowels can be arranged amongst themselves in 2 ! = 2 ways. 

∴ The total number of words = 2 × 120 = 240. 

6. Consonants occur together: Regarding these consonants as one letter the three letters E, I, (PNCL) can be arranged in 3 !, i.e., 6 ways. The letters PNCL can be arranged among themselves in 4 ! = 24 ways. 

∴ The number of words in which consonants occur together = 6 × 24 = 144. 

7. Vowels occupy even places. 

x x x x x x
1 2 3 4 5 6

There are 6 letters and 3 even places. E can be placed in any one of the three even places in 3P1, i.e., 3 ways. Having placed E in any one of these places, I can be placed in any one of the remaining 2 places in 2P1, i.e., 2 ways. Thus, the vowels can occupy even places in 2 × 3 = 6 ways. After the vowels have been placed the remaining 4 letters can take up their positions in 4P4, i.e., 24 ways. 

∴ The total number of words = 6 × 24 = 144.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...