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 :
Correct Option: , 4
Total number of ways placing 10 different balls
in 4 distinct boxes $=4^{10}$
Since, two of the 4 distinct boxes contains exactly 2
and 3 balls.
Then, there are three cases to place exactly 2 and 3
balls in 2 of the 4 boxes.
Case-1: When boxes contains balls in order $2,3,0,5$
Then, number of ways of placing the balls
$=\frac{10 !}{2 ! \times 3 ! \times 0 ! \times 5 !} \times 4 !$
Case-2: When boxes contains ball in order $2,3,1,4$.
Then, number of ways of placing the balls
$=\frac{10 !}{2 ! \times 3 ! \times 1 ! \times 4 !} \times 4 !$
Case-3: When boxes contains ball in order 2, 3, 2,3 Then, number of ways of placing the balls
$=\frac{10 !}{(2 !)^{2} \times(3 !)^{2} \times 2 ! \times 3 !} \times 4 !$
Therefore, number of ways of placing the balls that contains exactly 2 and 3 balls.
$=\frac{10 !}{2 ! \times 3 ! \times 0 ! \times 5 !} \times 4 !+\frac{10 !}{2 ! \times 3 ! \times 1 ! \times 4 !} \times 4 !$
$+\frac{10 !}{(2 !)^{2} \times 2 ! \times(3 !)^{2} \times 2 !} \times 4 !$
$=2^{5} \times 17 \times 945$
Hence, the required probability
$=\frac{2^{5} \times 17 \times 945}{4^{10}}=\frac{17 \times 945}{2^{15}}$
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