In how many ways can the letters of the word ‘STRANGE’ be arranged

1.	In how many ways can the letters of the word ‘STRANGE’ be arranged so that the vowels occupy only the odd places?
Answer» Given as The word ‘STRANGE’ Here's 7 letters in the word ‘STRANGE’, which includes 2 vowels (A,E) and 5 consonants (S,T,R,N,G). The vowels occupy only the odd places Here are 7 letters in the word ‘STRANGE’. Out of these letters (A,E) are the vowels. Here are 4 odd places in the word ‘STRANGE’. The two vowels can be arranged in ⁴P₂ ways. The remaining 5 consonants an be arranged among themselves in ⁵P₅ ways. Therefore, the total number of arrangements is On using the formula, P (n, r) = n!/(n – r)! P (4, 2) × P (5, 5) = 4!/(4 – 2)! × 5!/(5 – 5)! = 4!/2! × 5! = (4 × 3 × 2!)/2! × 5! = 4 × 3 × 5 × 4 × 3 × 2 × 1 = 12 × 120 = 1440 Thus, the number of arrangements therefore that the vowels occupy only odd positions is 1440.

In how many ways can the letters of the word ‘STRANGE’ be arranged so that the vowels occupy only the odd places?

Answer»

Given as

The word ‘STRANGE’

Here's 7 letters in the word ‘STRANGE’, which includes 2 vowels (A,E) and 5 consonants (S,T,R,N,G).

The vowels occupy only the odd places

Here are 7 letters in the word ‘STRANGE’. Out of these letters (A,E) are the vowels.

Here are 4 odd places in the word ‘STRANGE’. The two vowels can be arranged in ⁴P₂ ways.

The remaining 5 consonants an be arranged among themselves in ⁵P₅ ways.

Therefore, the total number of arrangements is

On using the formula,

P (n, r) = n!/(n – r)!

P (4, 2) × P (5, 5) = 4!/(4 – 2)! × 5!/(5 – 5)!

= 4!/2! × 5!

= (4 × 3 × 2!)/2! × 5!

= 4 × 3 × 5 × 4 × 3 × 2 × 1

= 12 × 120

= 1440

Thus, the number of arrangements therefore that the vowels occupy only odd positions is 1440.