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In an election no. of candidates exceeds the number to be elected by 2. If a man can vote in 56 ways, then the no. of candidates is

$\begin{array}{1 1} 5 \\ 6 \\ 7 \\ 8 \end{array}$

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  • $^nC_0+^nC_1+^nC_2+.................^nC_{n-2}+^nC_{n-1}+^nC_n=2^n$
Let the no. of candidates = $n$
$\therefore$ The no. to be elected= $n-2$
A man can vote for these candidates in
$^nC_1+^nC_2+^nC_3+..........^nC_{n-2}$ ways = $ 56 $(given)
$\Rightarrow\: 2^n-^nC_0-^nC_{n-1}-^nC_n=56$
$\Rightarrow\:2^n-1-n-1=56$
$\Rightarrow\:2^n=58+n$
$\Rightarrow\:n=6$
answered Aug 27, 2013 by rvidyagovindarajan_1
 

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