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14.Probability
hard
If $10$ different balls are to be placed in $4$ distinct boxes at random, then the probability that two of these boxes contain exactly $2$ and $3$ balls is
A
$\frac{945}{2^{11}}$
B
$\frac{965}{2^{11}}$
C
$\frac{945}{2^{10}}$
D
$\frac{965}{2^{10}}$
(JEE MAIN-2020)
Solution
Total ways $=4^{10}=\mathrm{n}$
Number of ways placing exactly 2 and 3 balls
in two of these boxes $=^{4} \mathrm{C}_{2} \times \frac{ 5!}{ 2! {3!}} \times 2! \times^{10} \mathrm{C}_{5} \times 2^{5}=\mathrm{m}$
Required probability $=\frac{\mathrm{m}}{\mathrm{n}}=\frac{945}{2^{10}}$
Standard 11
Mathematics
