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Worked Examples · Example 24

Q.Find the number of arrangements of the letters of the word FRAGILE. In how many of these arrangements

(i) do all the vowels occur together.
(ii) do all the vowels occur together and all the consonants occur together.
(iii) do all the vowels never occur together.
(iv) do the vowels occupy only even places.
(v) do the vowels occupy only odd places.
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FRAGILE has 7 distinct letters (3 vowels, 4 consonants); use the block method for "together" cases, subtraction for "never together", and direct placement-counting for the odd/even-place cases.

[!FORMULA]

Arrangements of nn distinct objects =n!=n!. Treating a group of kk objects as one block that must stay together: arrange the block with the rest as (n−k+1)!(n-k+1)! units, then arrange inside the block in k!k! ways — total (n−k+1)!×k!(n-k+1)!\times k!. "Never together" == Total −- "Together".

  1. Letters of FRAGILE: F,R,A,G,I,L,EF,R,A,G,I,L,E — 77 distinct letters. Vowels: A,I,EA,I,E (33 vowels). Consonants: F,R,G,LF,R,G,L (44 consonants).
  2. Total arrangements =7!=5040= 7! = 5040.
  3. (i) All vowels together: tie the 33 vowels into one block. Units to arrange: 44 consonants +1+ 1 vowel-block =5= 5 units ⇒5!\Rightarrow 5! ways; vowels inside the block can permute in 3!3! ways. Total =5!×3!=120×6=720= 5!\times3! = 120\times6 = 720.
  4. (ii) Vowels together AND consonants together: now there are 22 blocks (vowel-block, consonant-block); arrange the 22 blocks in 2!2! ways, vowels inside in 3!3! ways, consonants inside in 4!4! ways. Total =2!×3!×4!=2×6×24=288= 2!\times3!\times4! = 2\times6\times24 = 288.
  5. (iii) Vowels never all together: Total −- (vowels together) =5040−720=4320= 5040 - 720 = 4320. …

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