If 10 different balls are to be placed
Question:

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 :

1. (1) $\frac{965}{2^{11}}$

2. (2) $\frac{965}{2^{10}}$

3. (3) $\frac{945}{2^{10}}$

4. (4) None of these

Correct Option: , 4

Solution:

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

$=\frac{2^{5} \times 17 \times 945}{4^{10}}=\frac{17 \times 945}{2^{15}}$