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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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The number of arrangements of the letters of the word BHARAT taking 3 at a time is

$\begin{array}{1 1}(A)\;72\\(B)\;120\\(C)\;14\\(D)\;\text{None of these}\end{array} $

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
One pair one different arrangement can be$\rightarrow 1C_1\times 4C_1\times \large\frac{3!}{2!}$
$\Rightarrow 1\times 4\times 3=12$
Three different $\rightarrow 5C_3$
Total arrangements $\rightarrow 5C_3\times 3!$
$\Rightarrow \large\frac{5!}{3!2!}$$\times 3!$
$\Rightarrow \large\frac{5\times 4}{2}$$\times 6$
$\Rightarrow 60$
$\therefore$ The required no of ways =72
Hence (A) is the correct answer.
answered Jun 20, 2014 by sreemathi.v

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