Question -
Answer -
Given:
The word ‘STRANGE’
There are 7 letters inthe word ‘STRANGE’, which includes 2 vowels (A,E) and 5 consonants (S,T,R,N,G).
(i) the vowels cometogether?
Considering 2 vowelsas one letter so we will have 6 letters which can be arranged in 6P6 ways.
(A,E) can be puttogether in 2P2 ways.
Hence, the requirednumber of words are
By using the formula,
P (n, r) = n!/(n – r)!
P (6, 6) × P (2, 2) =6!/(6 – 6)! × 2!/(2 – 2)!
= 6! × 2!
= 6 × 5 × 4 × 3 × 2 ×1 × 2 × 1
= 720 × 2
= 1440
Hence, total number ofarrangements in which vowels come together is 1440.
(ii) the vowels never cometogether?
The total number ofletters in the word ‘STRANGE’ is 7P7 = 7! = 7 ×6 × 5 × 4 × 3 × 2 × 1 = 5040
So,
Total number of wordsin which vowels never come together = total number of words – number of wordsin which vowels are always together
= 5040 – 1440
= 3600
Hence, the totalnumber of arrangements in which vowel never come together is 3600.
(iii) the vowels occupy onlythe odd places?
There are 7 letters inthe word ‘STRANGE’. Out of these letters (A,E) are the vowels.
There are 4 odd placesin the word ‘STRANGE’. The two vowels can be arranged in 4P2 ways.
The remaining 5consonants an be arranged among themselves in 5P5 ways.
So, the total numberof arrangements is
By 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
Hence, the number ofarrangements so that the vowels occupy only odd positions is 1440.