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In How Many Ways Can the Letters of the Word 'Failure' Be Arranged So that the Consonants May Occupy Only Odd Positions?

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प्रश्न

In how many ways can the letters of the word 'FAILURE' be arranged so that the consonants may occupy only odd positions?

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उत्तर

There are 7 letters in the word FAILURE.
We wish to find the total number of arrangements of these 7 letters so that the consonants occupy only odd positions.
There are 3 consonants and 4 odd positions. These 3 consonants can be arranged in the 4 positions in 4! ways.
Now, the remaining 4 vowels can be arranged in the remaining 4 positions in 4! ways.

By fundamental principle of counting:
Total number of arrangements = 4! \[\times\] 4! = 576

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Factorial N (N!) Permutations and Combinations
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 16: Permutations - Exercise 16.4 [पृष्ठ ३६]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 16 Permutations
Exercise 16.4 | Q 1 | पृष्ठ ३६

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