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CBSE XI
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Permutations and Combinations
The number of triangles which can be formed by using the vertices of a regular polygon of (n+3) sides is 220.Then n is equal to
$\begin{array}{1 1}(A)\;8\\(B)\;9\\(C)\;10\\(D)\;11\end{array} $
cbse
math
class11
ch7
permutations-and-combinations
additionalproblem
q46
sec-b
medium
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asked
Jun 20, 2014
by
sreemathi.v
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1 Answer
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$C(n,r)=\large\frac{n!}{r!(n-r)!}$
Number of triangles =$n+3 C_3=220$
$\large\frac{(n+3)(n+2)(n+1)}{3!}=$$220$
$(n+1)(n+2)(n+3)=1320$
$\Rightarrow 12\times 10\times 11$
$\Rightarrow (9+1)(9+2)(9+3)$
$n=9$
Hence (B) is the correct answer.
answered
Jun 20, 2014
by
sreemathi.v
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