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Worked Examples · Example 26

Q.Find the number of permutations of the letters of the word ENGINEERING. In how many of these arrangements:

(i) do the words begin with E and end with G
(ii) do all the vowels come together
(iii) do all the vowels never come together
(iv) no two vowels come together
Sikkim CbseNCERTSubjective· 5mImportance★★★★★est
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Use the repeated-letters permutation formula for the total, fix-and-arrange for the begin/end case, the block method for "vowels together", subtraction for "never together", and the gap method for "no two vowels adjacent".

[!FORMULA]

Permutations of nn letters with repetitions p1,p2,…p_1,p_2,\dots each =n!p1! p2!⋯= \dfrac{n!}{p_1!\,p_2!\cdots}. "No two together" (gap method): arrange the other objects first, then place the restricted objects into the gaps created (including the two ends) so no two of them are adjacent.

  1. Letters of ENGINEERING: E,N,G,I,N,E,E,R,I,N,GE,N,G,I,N,E,E,R,I,N,G — total n=11n=11 letters. Counting occurrences: E=3E=3, N=3N=3, G=2G=2, I=2I=2, R=1R=1 (check: 3+3+2+2+1=113+3+2+2+1=11 ✓).
  2. Vowels present: EE (33 times), II (22 times) ⇒\Rightarrow 55 vowels total. Consonants: NN (33), GG (22), RR (11) ⇒\Rightarrow 66 consonants total.
  3. Total arrangements =11!3! 3! 2! 2!=399168006×6×2×2=39916800144=277200= \dfrac{11!}{3!\,3!\,2!\,2!} = \dfrac{39916800}{6\times6\times2\times2} = \dfrac{39916800}{144} = 277200.
  4. (i) Begin with E and end with G: fix one E at the start and one G at the end. Remaining letters to arrange in the middle 99 positions: E=2,N=3,G=1,I=2,R=1E=2, N=3, G=1, I=2, R=1 (total 99). Arrangements =9!2! 3! 1! 2! 1!=3628802×6×1×2×1=36288024=15120= \dfrac{9!}{2!\,3!\,1!\,2!\,1!} = \dfrac{362880}{2\times6\times1\times2\times1} = \dfrac{362880}{24} = 15120.
  5. (ii) All vowels come together: tie the 55 vowels (E=3,I=2E{=}3,I{=}2) into one block. Units to arrange: 66 consonants +1+ 1 vowel-block =7=7 units, with consonant repeats N=3,G=2,R=1N{=}3,G{=}2,R{=}1: 7!3! 2! 1!=504012=420\dfrac{7!}{3!\,2!\,1!} = \dfrac{5040}{12}=420 ways. Inside the block, the 55 vowels permute in 5!3! 2!=12012=10\dfrac{5!}{3!\,2!} = \dfrac{120}{12}=10 ways. Total =420×10=4200= 420\times10 = 4200. …

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