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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations

There are two urns.Urn A has 3 distinct red balls and urn B has 9 distinct blue balls.From each urn two balls are taken out at random and then transferred to the other.The number of ways in which this can be done is

$\begin{array}{1 1}(A)\;3\\(B)\;36\\(C)\;66\\(D)\;108\end{array} $

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1 Answer

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
Total no. of balls in urn A=3
Total no. of balls in urn B=9
The number of ways in which two balls from urn A and two balls from urn B can be selected =$3C_2\times 9C_2$
$3C_2=\large\frac{3!}{2!1!}$$=3$
$9C_2=\large\frac{9!}{2!7!}=\frac{9\times 8\times 7!}{2!\times 7!}$$=36$
$3C_2\times 9C_2=3\times 36$
$\Rightarrow 108$
Hence (D) is the correct answer.
answered Jun 20, 2014 by sreemathi.v
 

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