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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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The number of words of four letters containing equal number of vowels and consonants ,repetition allowed is

$\begin{array}{1 1}(A)\;(105)^2\\(B)\;210\times 243\\(C)\;105\times 243\\(D)\;\text{None of these}\end{array} $

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
Total vowel=5
Total consonants=21
$\therefore$ The number of selection of 1 pair of vowels and 1 pair of consonants =$5C_1\times 21C_1$
The required number of words =$5C_1\times 21C_1\times \large\frac{4!}{2!2!}$$+5C_2\times 21C_2\times 41$
$\Rightarrow 5\times 21 \times \large\frac{4\times 3\times 2\times 1}{2\times 2}$$+\large\frac{5!}{2!3!}\times \frac{21!}{2!19!}$$\times 4!$
$\Rightarrow 105\times 6+10(210\times 24)$
$\Rightarrow 105\times 6+10(5040)$
$\Rightarrow 630+50400$
$\Rightarrow 57030$
Hence (D) is the correct answer.
answered May 19, 2014 by sreemathi.v

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