Find the number of words with or without 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?

Solution

Total letters in word 'AGAIN' (A → 2, G → 1, I → 1, N → 1) = 5
Number of words that begin with A = $space space fraction numerator straight P presuperscript 2 subscript 2 over denominator 2 factorial end fraction cross times straight P presuperscript 4 subscript 4 space equals space 1 cross times 4 factorial space equals space 24$
(Fix A at first place, so that the remaining 4 letters are different)
Number of words with begin with G = $straight P presuperscript 1 subscript 1 space cross times space fraction numerator straight P presuperscript 4 subscript 4 over denominator 2 factorial end fraction space equals space 1 cross times fraction numerator 4 factorial over denominator 2 factorial end fraction equals 24 over 12 equals 12$
Number of words that begin with I = $straight P presuperscript 1 subscript 1 cross times fraction numerator straight P presuperscript 4 subscript 4 over denominator 2 factorial end fraction equals 12$
Thus, so far we have formed 24 + 12 + 12 = 48 words.
The 49th word will begin with N and it is NAAGI and the 50th word is NAAIG.

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Permutations And Combinations

Find the number of words with or without 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?

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