Permutations And Combinations

Question
CBSEENMA11014165

In how many ways can the letters of word ‘ASSASSINATION’ be arranged so that the arrangements be such that they start with O and end with T and the S’s are all together ? 

Solution

Number of letters:
                A → 3, S → 4
                 I → 2, N → 2
                 T → 1, O → 1
Total letters 13.
Tie the four S's.
Number of arrangements = fraction numerator straight P presuperscript 4 subscript 4 over denominator 4 factorial end fraction space equals space fraction numerator 4 factorial over denominator 4 factorial end fraction equals 1                           ...(i)
Mix with remaining to give:
[3 (A) + 2 (I) + 2 (N) + 1 (T) + 1 (O)] + 1 = 10
Fix O at first place.
Number of permutations = straight P presuperscript 1 subscript 1 equals 1                                        ...(ii)
Fix T at the last place. 
Number of permutations = straight P presuperscript 1 subscript 1 equals 1                                        ...(iii)
Now, we have
3 (A) + 2 (I) + 2 (N) + 1 (four S's tied) = 8 letters
Number of arrangements = fraction numerator 8 factorial over denominator 3 factorial space 2 factorial space 2 factorial end fraction space equals space fraction numerator 8 cross times 7 cross times 6 cross times 5 cross times 4 cross times 3 cross times 2 cross times 1 over denominator 3 cross times 2 cross times 1 cross times 2 cross times 1 cross times 2 cross times 1 end fraction equals 1680.
Hence, by fundamental principle of counting, the total number of permutations
                                        = 1 x 1 x 1 x 1680 = 1680.
  



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