Given as

The word ‘GANESHPURI’

Here 10 letters in the word ‘GANESHPURI’. Total number of words formed is ¹⁰P₁₀ = 10!

(i) the letter G always occupies the first place?

If we fix the first position with letter G, now remaining number of letters is 9.

Number of arrangements of 9 things, taken all at a time is ⁹P₉ = 9! Ways.

Thus, a possible number of words using letters of ‘GANESHPURI’ starting with ‘G’ is 9!

(ii) the letter P and I respectively occupy the first and last place?

If we fix the first position with letter P and I in the end, now remaining number of letters is 8.

Number of arrangements of 8 things, taken all at a time is ⁸P₈ = 8! Ways.

Thus, a possible number of words using letters of ‘GANESHPURI’ starting with ‘P’ and ending with ‘I’ is 8!

(iii) Are the vowels always together?

Here 4 vowels and 6 consonants in the word ‘GANESHPURI’.

Considering 4 (A,E,I,U) vowels as one letter, now total number of letters is 7 (A,E,I,U, G, N, S, H , P, R)

Number of arrangements of 7 things, taken all at a time is ⁷P₇ = 7! Ways.

(A, E, I, U) can be put together in 4! Ways.

Thus, total number of arrangements in which vowels come together is 7! × 4!

(iv) the vowels always occupy even places?

The number of vowels in the word ‘GANESHPURI’ = 4(A, E, I, U)

The number of consonants = 6(G, N, S, H, R, I)

The even positions are 2, 4, 6, 8 or 10

Then, we have to arrange 10 letters in a row such that vowels occupy even places. Here 5 even places (2, 4, 6, 8 or 10). 4 vowels can be arranged in these 5 even places in ⁵P₄ ways.

The remaining 5 odd places (1, 3, 5, 7, 9) are to be occupied by the 6 consonants in ⁶P₅ ways.

Therefore, by using the formula,

P (n, r) = n!/(n – r)!

P (5, 4) × P (6, 5) = 5!/(5 – 4)! × 6!/(6 – 5)!

= 5! × 6!

Thus, number of arrangements therefore that the vowels occupy only even positions is 5! × 6!