Question -
Answer -
Given:
The word ‘GANESHPURI’
There are 10 lettersin the word ‘GANESHPURI’. The total number of words formed is 10P10 =10!
(i) the letter G alwaysoccupies the first place?
If we fix the firstposition with letter G, then remaining number of letters is 9.
The number ofarrangements of 9 things, taken all at a time is 9P9 =9! Ways.
Hence, a possiblenumber of words using letters of ‘GANESHPURI’ starting with ‘G’ is 9!
(ii) the letter P and Irespectively occupy the first and last place?
If we fix the firstposition with letter P and I in the end, then remaining number of letters is 8.
The number ofarrangements of 8 things, taken all at a time is 8P8 =8! Ways.
Hence, a possiblenumber of words using letters of ‘GANESHPURI’ starting with ‘P’ and ending with‘I’ is 8!
(iii) Are thevowels always together?
There are 4 vowels and6 consonants in the word ‘GANESHPURI’.
Consider 4 (A,E,I,U)vowels as one letter, then total number of letters is 7 (A,E,I,U, G, N, S, H ,P, R)
The number ofarrangements of 7 things, taken all at a time is 7P7 =7! Ways.
(A, E, I, U) can beput together in 4! Ways.
Hence, total number ofarrangements in which vowels come together is 7! × 4!
(iv) the vowels alwaysoccupy even places?
Number of vowels inthe word ‘GANESHPURI’ = 4(A, E, I, U)
Number of consonants =6(G, N, S, H, R, I)
Even positions are 2,4, 6, 8 or 10
Now, we have toarrange 10 letters in a row such that vowels occupy even places. There are 5even places (2, 4, 6, 8 or 10). 4 vowels can be arranged in these 5 even placesin 5P4 ways.
Remaining 5 odd places(1, 3, 5, 7, 9) are to be occupied by the 6 consonants in 6P5 ways.
So, by using theformula,
P (n, r) = n!/(n – r)!
P (5, 4) × P (6, 5) =5!/(5 – 4)! × 6!/(6 – 5)!
= 5! × 6!
Hence, number ofarrangements so that the vowels occupy only even positions is 5! × 6!