Example 22 - Chapter 6 Class 11 Permutations and Combinations
Last updated at April 16, 2024 by Teachoo
Examples
Example 1
Example 2
Example 3 Important
Example 4 Important
Example 5
Example 6 (i)
Example 6 (ii)
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Example 8 Important
Example 9 Important
Example 10
Example 11 Important
Example 12 (i) Important
Example 12 (ii)
Example 13
Example 14 Important
Example 15
Example 16 Important
Example 17
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Example 19 Important
Example 20 Important
Example 21 Important
Example 22 Important You are here
Example 23 Important
Example 24 Important
Chapter 6 Class 11 Permutations and Combinations
Serial order wise
- Ex 6.1
- Ex 6.2
- Ex 6.3
- Ex 6.4
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Examples
- Miscellaneous
Transcript
Example 22 Find the number of words with or without meaning which can be made using all the letters of the word AGAIN. If these words are written as in a dictionary, what will be the 50th word? ‘AGAIN’ = 2A, 1G, 1I & 1N In dictionary, letters appear alphabetically, 4 letters in which A, G, I, N Since, we arrange 4 letters, Number of words = 4P4 = 4! = 24 4 letters in which 2A, I, N Since, letters are repeating, number of words = 𝑛!/𝑝1!𝑝2!𝑝3! Number of letter = n = 4 Since 2A, p1 = 4 Number of words = 4!/2! = 12 4 letters in which 2A, G, N Since, letters are repeating, number of words = 𝑛!/𝑝1!𝑝2!𝑝3! Number of letter = n = 4 Since 2A, p1 = 4 Number of words = 4!/2! = 12 Thus, Total no of words starting with A, G, & I = 24 + 12 + 12 = 48 Hence, 49th word will be start from N i.e. N A A G I & remaining four rearrange according to a dictionary Thus, The 50th word is N A A I G Hence the 50th word will be NAAIG
