How many different words can be formed from the letters of the word GANESHPURI'? In how many of these words:
(i) the letter G always occupies the. first place?
(ii) the letters P and 1 respectively occupy first and last place ?
(iii) the vowels are always together ?
(iv) the vowels always occupy even places?
How many different words can be formed from the letters of the word GANESHPURI'? In how many of these words:
(i) the letter G always occupies the. first place?
(ii) the letters P and 1 respectively occupy first and last place ?
(iii) the vowels are always together ?
(iv) the vowels always occupy even places?
There are 10 letters in the word 'GANESHPURI'. The total number of words formed is equal to 10P10=10!
(i) If we fix up G in the beginning, then the remaining 9 letters can be arranged in 9P9=9! ways
(ii) If we fix up P in the begining and I at the end, begining 8 letters can be arranged in 8P8=8!
(iii) There are 4 vowels and 6 consonants the word 'GANESHPURI'. Considening 4 vowels as one letter, We have 7 letters which can be arranged in 7p77! ways.
A, E, U, I can be put together in 4! ways. Hence, required number of words=7!×4!
(iv) We have to arrange 10 letters in a row such that vowels occupy even places. There are 5 even places (2, 4, 6, 8, 10). 4 vowels 531 can be arranged in these 5 even places in 5p4 ways.
Remaining 5 odds places (1,3,5,7,9) are to occupied by the 6 consonants.
This can be done in 6P5 ways.
Hence, the total number of words in which vowels occupy even places=5P4×6P5
= 5!(5−4)!×6!(6−1)! = 5!×6!
There are 10 letters in the word 'GANESHPURI'. The total number of words formed is equal to 10P10=10!
(i) If we fix up G in the beginning, then the remaining 9 letters can be arranged in 9P9=9! ways
(ii) If we fix up P in the begining and I at the end, begining 8 letters can be arranged in 8P8=8!
(iii) There are 4 vowels and 6 consonants the word 'GANESHPURI'. Considening 4 vowels as one letter, We have 7 letters which can be arranged in 7p77! ways.
A, E, U, I can be put together in 4! ways. Hence, required number of words=7!×4!
(iv) We have to arrange 10 letters in a row such that vowels occupy even places. There are 5 even places (2, 4, 6, 8, 10). 4 vowels 531 can be arranged in these 5 even places in 5p4 ways.
Remaining 5 odds places (1,3,5,7,9) are to occupied by the 6 consonants.
This can be done in 6P5 ways.
Hence, the total number of words in which vowels occupy even places=5P4×6P5
= 5!(5−4)!×6!(6−1)! = 5!×6!
