Permutations And Combinations

Question
CBSEENMA11014169

How many different words can be formed by using the letters of the word ‘ALLAHABAD?

(a) In how many of these do the vowels occupy even positions.

(b) In how many of these, the two L’s do not come together?

Solution

Number of letters in word 'ALLAHABAD' = (A → 4, L → 2, H → 1, B → 1, D → 1) = 9
Number of arrangements   = fraction numerator straight n factorial over denominator straight p factorial space straight q factorial end fraction space equals space fraction numerator 9 factorial over denominator 4 factorial space 2 factorial end fraction space equals space fraction numerator 9 cross times 8 cross times 7 cross times 6 cross times 5 over denominator 2 end fraction equals 7560.
(a) There are only four A's as vowels.
          
They can occupy even places (2, 4, 6, 8) in space space fraction numerator straight P presuperscript 4 subscript 4 over denominator 4 factorial end fraction  ways
∴   Number of ways in which vowels occupying even places = 1
We are left with 5 places and letters (L → 2, H → 1, B → 1, D → 1).
Number of permutations = fraction numerator 5 factorial over denominator 2 factorial end fraction space equals space 5 cross times 4 cross times 3 equals 60.
Hence, total number of arrangements in which A's occupy even places
                            = 1 x 60 = 60.
(b) We first find the number of arrangements in which two L's are not together:
      Number of arrangements in which two L's are together
                 equals fraction numerator straight P presuperscript 2 subscript 2 over denominator 2 factorial end fraction cross times fraction numerator straight P presuperscript left parenthesis 7 plus 1 right parenthesis end presuperscript subscript 7 plus 1 end subscript over denominator 4 factorial end fraction space equals space 1 cross times fraction numerator 8 factorial over denominator 4 factorial end fraction space equals space 8 cross times 7 cross times 6 cross times 5 space equals space 1680.
Hence, the number of arrangements in which the two L's are not together
     = (Total arrangements) - (the number of arrangements in which the two L's are together)
     = 7560 - 1680 = 5880.

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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.