Question 14 Marks
Five balls are to be placed in three boxes, where each box can contain upto five balls. Find the number of ways if no box is to remain empty.
Answer
View full question & answer→Let boxes be named as I, II, III
Let sets A, B, C represent cases in which boxes I, II, III remain empty
Then $A \cup B \cup C$ represent the cases in which at least one box remains empty.
Then we use method of indirect counting
Required number $=$ Total number of distributions $-n(A \cup B \cup C)$
$n(A \cup B \cup C)$ represent the number of undesirable cases
Total number of distributions $=3 \times 3 \times 3 \times 3 \times 3=3^5=243$
$
n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A \cap B)-n(B \cap C)-n(C \cap A)+n(A \cap B
$
$\cap \mathrm{C})$
(iii)
In box I is empty then every ball has two places (boxes) to go.
Similarly for box II and III.
$
\therefore \mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})+\mathrm{n}(\mathrm{C})=3 \times 2^5
$
If boxes I and II remain empty then all balls go to box III
Similarly we would have two more cases.
$
\therefore \mathrm{n}(\mathrm{A} \cap \mathrm{B})+\mathrm{n}(\mathrm{B} \cap \mathrm{C})+\mathrm{n}(\mathrm{C} \cap \mathrm{A})=3 \times 1^5 \ldots \ldots(\mathrm{v})
$
$\therefore \mathrm{n}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0$
(vi) [as all boxes cannot be empty]
Substitute from (iv), (v), (vi) to (iii) to get
$
\begin{aligned}
& n(A \cup B \cup C)=3 \times 2^5-3 \times 1^5 \\
& =96-3 \\
& =93
\end{aligned}
$
Substitute $n(A \cup B \cup C)$ and from (ii) to (i), we get
Required number $=243-93=150$
Let sets A, B, C represent cases in which boxes I, II, III remain empty
Then $A \cup B \cup C$ represent the cases in which at least one box remains empty.
Then we use method of indirect counting
Required number $=$ Total number of distributions $-n(A \cup B \cup C)$
$n(A \cup B \cup C)$ represent the number of undesirable cases
Total number of distributions $=3 \times 3 \times 3 \times 3 \times 3=3^5=243$
$
n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A \cap B)-n(B \cap C)-n(C \cap A)+n(A \cap B
$
$\cap \mathrm{C})$
(iii)
In box I is empty then every ball has two places (boxes) to go.
Similarly for box II and III.
$
\therefore \mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})+\mathrm{n}(\mathrm{C})=3 \times 2^5
$
If boxes I and II remain empty then all balls go to box III
Similarly we would have two more cases.
$
\therefore \mathrm{n}(\mathrm{A} \cap \mathrm{B})+\mathrm{n}(\mathrm{B} \cap \mathrm{C})+\mathrm{n}(\mathrm{C} \cap \mathrm{A})=3 \times 1^5 \ldots \ldots(\mathrm{v})
$
$\therefore \mathrm{n}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0$
(vi) [as all boxes cannot be empty]
Substitute from (iv), (v), (vi) to (iii) to get
$
\begin{aligned}
& n(A \cup B \cup C)=3 \times 2^5-3 \times 1^5 \\
& =96-3 \\
& =93
\end{aligned}
$
Substitute $n(A \cup B \cup C)$ and from (ii) to (i), we get
Required number $=243-93=150$