How many different words can be formed from the letters of the word ‘GANESHPURI’? In how many of these words

Given:

The word ‘GANESHPURI’

There are 10 letters in the word ‘GANESHPURI’. The 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, then remaining number of letters is 9.

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

Hence, 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, then remaining number of letters is 8.

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

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

(iii) Are the vowels always together?

There are 4 vowels and 6 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 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.

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

(iv) the vowels always occupy even places?

Number of vowels in the 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 to arrange 10 letters in a row such that vowels occupy even places. There are 5 even places (2, 4, 6, 8 or 10). 4 vowels can be arranged in these 5 even places in ⁵P₄ ways.

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

P (5, 4) × P (6, 5)

=\(\dfrac{ 5!}{(5 – 4)!} × \dfrac{6!}{(6 – 5)!}\)

= 5! × 6!

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