Permutations And Combinations

Question
CBSEENMA11014167

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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Some More Questions From Permutations and Combinations Chapter

Determine K, so that K + 2, 4K – 6 and 3K – 2 are three consecutive terms of an A.P.