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Permutations and Combinations question

2025 · 23 Jan · Shift 1 · Q28
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Permutations and Combinations question

2025 · 23 Jan · Shift 1 · Q28

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
The number of words, which can be formed using all the letters of the word "DAUGHTER", so that all the vowels never come together, is :
  1. A
    34000
  2. B
    37000
  3. C
    35000
  4. D
    36000
View written solutionFree

Correct answer: D

  1. Identify the letters and vowels

The word DAUGHTER has 8 distinct letters:

D,A,U,G,H,T,E,RD, A, U, G, H, T, E, RD,A,U,G,H,T,E,R

Vowels are:

A,U,EA, U, EA,U,E

So:

  • Number of vowels =3=3=3
  • Number of consonants =5=5=5 namely D,G,H,T,RD,G,H,T,RD,G,H,T,R
  1. Total number of arrangements of all 8 letters

Since all letters are distinct, total arrangements are:

8!=403208! = 403208!=40320

  1. Count arrangements where all vowels come together

If all 3 vowels are together, treat them as one block.

Then the objects are:

  • 1 vowel block (AUE)(AUE)(AUE)
  • 5 consonants

So total objects =6=6=6.

These 6 objects can be arranged in:

6!6!6!

Now, within the vowel block, the 3 vowels can be arranged among themselves in:

3!3!3!

Hence, number of arrangements with all vowels together:

6!×3!=720×6=43206! \times 3! = 720 \times 6 = 43206!×3!=720×6=4320

  1. Count arrangements where all vowels never come together

This means we exclude the cases where all vowels are together:

8!−6!⋅3!8! - 6!\cdot 3!8!−6!⋅3!

=40320−4320= 40320 - 4320=40320−4320

=36000= 36000=36000

  1. Match with options

360003600036000 corresponds to Option D.

  1. Comparison with stored answer

Stored correct answer is D, and our derived answer is also D.

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