मराठी

The Letters of the Word 'Zenith' Are Written in All Possible Orders. How Many Words Are Possible If All These Words Are Written Out as in a Dictionary? What is the Rank of the Word 'Zenith'? - Mathematics

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प्रश्न

The letters of the word 'ZENITH' are written in all possible orders. How many words are possible if all these words are written out as in a dictionary? What is the rank of the word 'ZENITH'?

उत्तर

In a dictionary, the words are arranged in alphabetical order. Therefore, in the given problem, we must consider the words beginning with E, H, I, N, T and Z.

∴ Number of words starting with  E = 5! = 120
    Number of words starting with  H = 5! = 120
    Number of words starting with  I = 5! = 120
    Number of words starting with  N = 5! = 120
    Number of words starting with  T = 5! = 120
Now, the word will start with the letter Z.
After Z, alphabetically, the next letter would be E, which is as per the requirement of the word ZENITH.
After ZE, alphabetically, the next letter would be H, i.e. ZEH. The remaining three letters can be arranged in 3! ways.
Now, the next letter would be I, i.e. ZEI. The remaining three letters can be arranged in 3! ways.
Now, the next letter would be N, which as per the requirement of the word ZENITH.

After ZEN, alphabetically, the next letter would be H, i.e. ZENH. The remaining two letters can be arranged in 2! ways.
The next letter would now be I, i.e. ZENI, which is as per the requirement of the word ZENITH.
H will come after ZENI, which would be followed by T.
The word formed is ZENIHT.
The next word would be ZENITH.
 Total number of intermediate words = 5\[\times\]120 + 3! + 3! + 2! + 1 + 1 = 616

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Factorial N (N!) Permutations and Combinations
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 16: Permutations - Exercise 16.5 [पृष्ठ ४४]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 16 Permutations
Exercise 16.5 | Q 28 | पृष्ठ ४४

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