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Permutations and Combinations question

2021 · 31 Aug · Shift 1 · Q42
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Permutations and Combinations question

2021 · 31 Aug · Shift 1 · Q42

JEE MainMathematicsPermutations and CombinationsNumerical+4 / −1
The number of six letter words (with or without meaning), formed using all the letters of the word 'VOWELS', so that all the consonants never come together, is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 576

  1. Identify vowels and consonants in the word VOWELS.

    The letters are: V,O,W,E,L,SV, O, W, E, L, SV,O,W,E,L,S.

    • Vowels: O,EO, EO,E
    • Consonants: V,W,L,SV, W, L, SV,W,L,S

    All 666 letters are distinct.

  2. Total number of 6-letter arrangements using all letters:

    6!=7206! = 7206!=720

  3. We need arrangements such that all the consonants never come together.
    So first count arrangements where all 4 consonants are together, then subtract from total.

  4. Count arrangements where all consonants are together:

    Treat the 4 consonants V,W,L,SV, W, L, SV,W,L,S as one block.

    Then we have these 3 objects:

    • (VWLS)(VWLS)(VWLS) block
    • OOO
    • EEE

    These 3 objects can be arranged in:

    3!3!3!

    ways.

    Inside the consonant block, the 4 consonants can be arranged in:

    4!4!4!

    ways.

    Therefore, total arrangements with all consonants together:

    3!⋅4!=6⋅24=1443! \cdot 4! = 6 \cdot 24 = 1443!⋅4!=6⋅24=144

  5. Required number:

    6!−3!⋅4!=720−144=5766! - 3!\cdot 4! = 720 - 144 = 5766!−3!⋅4!=720−144=576

  6. Final answer:

    576\boxed{576}576​

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