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Permutations — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSPermutations4 Marks
Question
How many different words can be formed from the letters of the word 'GANESHPURI'? In how many of these words:
  1. The letter G always occupies the first place?
  2. The letters P and I respectively occupy first and last place?
  3. The vowels are always together?
  4. The vowels always occupy even places?
✓

Answer

There are 10 letters in the word 'GANESHPURI'. The total number of words formed is equal to $^{10}P_{10} = 10!$
  1. If we fix up G in the begining, then the remaining 9 letters can be arranged in $^9P_9 = 9!$ ways
  2. If we fix up P in the begining and I at the end, begining 8 letters can be arranged in $^8P_8 = 8!$.
  3. There are 4 vowels and 6 consonants in the word 'GANESHPURI'.
Considening 4 vowels as one letter,
We have 7 letters which can be arranged in $^7P_7 = 7!$ ways.
A, E, U, I can be put together in 4! ways.
Hence, required number of words = 7! × 4!.
  1. We have to arrange 10 letters in a row such that vowels occupy even places. There are 5 even places (2, 4, 6, 8, 10). 4 vowels can be arranged in these 5 even places in $^5p_4$ ways.
Remaining 5 odd places (1, 3 , 5, 7, 9) are to be occupied by the 6 consonants.
This can be done in $^6C_5$​​​​​​​ ways.
Hence, the total number of words in which vowels occupy even places $=\ ^5\text{P}_4 \times \ ^6\text{P}_5$
$=\frac{5!}{(5-4)!}\times \frac{6!}{(6-1)!}$
$=5!\times 6!$

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