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6.Permutation and Combination
hard
There are two urns. Urm $A$ has $3$ distinct red balls and urn $B$ has $9$ distinct blue balls. From each urm two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is
A
$36$
B
$66$
C
$108$
D
$3$
(AIEEE-2010)
Solution
Total no. of balls in urn $\mathrm{A}=3$
Total no. of balls in urn $\mathrm{B}=9$
The number of ways in which two balls from urn $A$ and two balls from urn $B$ can be
selected $=3 C_{2} \times 9 C_{2}$
$3 C_{2}=\frac{3 !}{2 ! 1 !}=3$
$9 C_{2}=\frac{9 !}{2 ! 7 !}=\frac{9 \times 8 \times 7 !}{2 ! \times 7 !}=36$
$3 C_{2} \times 9 C_{2}=3 \times 36$
$\Rightarrow 108$
Standard 11
Mathematics
