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If 5 distinct balls are placed at random into 5 cells, then the probability that exactly one cell remains empty, is
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$48 / 125$
Clearly. 5 distinct balls can be placed into 5 olls in $5^{5}$ ways. Now, the number of ways of selecting one cell ${ }^{5} C_{1}$
Let the selected cell be empty, Now, we are left with 5 balls and 4 cells. Now, number of ways of placing 5 distinct balls into 4 cells such that each cell have atleast one $=4^{5}-{ }^{4} C_{1}(4-1)^{5}+{ }^{4} C_{2}(4-2)^{5}-{ }^{4} C_{3}(4-3)^{6}$
$=1024-4 \times 3^{5}+6 \times 2^{5}-4$
$=1024-972+192-4=1216-976$
$=240$
Thus, the number of ways of placing 5 distinct balls such that exactly one cell remains empty $={ }^{5} C_{1} \times 240$
$\therefore$ Required probability $=\frac{{ }^{5} C_{1} \times 240}{5^{5}}=\frac{240}{625}=\frac{48}{125}$
Let the selected cell be empty, Now, we are left with 5 balls and 4 cells. Now, number of ways of placing 5 distinct balls into 4 cells such that each cell have atleast one $=4^{5}-{ }^{4} C_{1}(4-1)^{5}+{ }^{4} C_{2}(4-2)^{5}-{ }^{4} C_{3}(4-3)^{6}$
$=1024-4 \times 3^{5}+6 \times 2^{5}-4$
$=1024-972+192-4=1216-976$
$=240$
Thus, the number of ways of placing 5 distinct balls such that exactly one cell remains empty $={ }^{5} C_{1} \times 240$
$\therefore$ Required probability $=\frac{{ }^{5} C_{1} \times 240}{5^{5}}=\frac{240}{625}=\frac{48}{125}$
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