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Olympiad workbook
Exam
Olympiad workbook
Subject
IOQM
Chapter
Permutation and Combination

Olympiad workbook · IOQM · Permutation and Combination

We will say that a rearrangement of the letters of a word has no...

We will say that a rearrangement of the letters of a word has no fixed letters if. When the rearrangement is placed directly below the word, no column has the same letter repeated. For instance, HBRATA is a rearrangement with no fixed letters of BHARAT. How many distinguishable rearrangement with no fixed letters does BHARAThave? (The two A's are considered identical)

Correct answer

84

Step-by-step solution

Let use assume the 2A′2 A^{\prime} 's to A1A_1 and A2A_2 BHA1RA2 T\mathrm{BHA}_1 \mathrm{RA}_2 \mathrm{~T} Number of rearrangements of these 6. =\left\lfloor 6\left(\frac{1}{\lfloor 2}-\frac{1}{\lfloor 3}+\frac{1}{\lfloor 4}-\frac{1}{\lfloor 5}+\frac{1}{\lfloor 6}\right)=265\right. Let PP be the set when A2A_2 occupies place of A1A_1 and θ\theta be the set when A1A_1 occupies place of A2A_2 n(P)=53,n(Q)=53n(n∩Q)=9\begin{array}{ll} n(P)=53, & n(Q)=53 \\ n(n \cap Q)=9 & \end{array} Hence required arrangements =12(265−n(P∪Q))=12(265−106+9)=1682=84\begin{aligned} & =\frac{1}{2}(265-n(P \cup Q)) \\ & =\frac{1}{2}(265-106+9) \\ & =\frac{168}{2}=84 \end{aligned}
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