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.
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.
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.