# Five balls are to be placed in three boxes, where each box can contain up to five balls. Find the number of ways if no box is to remain empty. - Mathematics and Statistics

Sum

Five balls are to be placed in three boxes, where each box can contain up to five balls. Find the number of ways if no box is to remain empty.

#### Solution

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 ∪ B ∪ C represents the cases in which at least one box remains empty.
Then we use the method of indirect counting
Required number = Total number of distributions – n(A ∪ B ∪ C) ...(i)
n(A ∪ B ∪ C) represent the number of undesirable cases
Total number of distributions
= 3 × 3 × 3 × 3 × 3
= 35
= 243 .......(ii)
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) n n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C) ...........(iii)
In box I is empty then every ball has two places (boxes) to go.
Similarly for box II and III.
∴ n(A) + n(B) + n(C) = 3 × 25 .....… (iv)
If boxes I and II remain empty
then all balls go to box III
Similarly we would have two more cases.
∴ n(A ∩ B) + n(B ∩ C) + n(C ∩ A)
= 3 × 15 ..........(v)
n(A ∩ B ∩ C) = 0 .......(vi) .........[("as all boxes"),("cannot be empty")]
Substitute from (iv), (v), (vi) to (iii) to get
n(A ∪ B ∪ C) = 3 × 25 – 3 × 15
= 96 – 3
= 93
Substitute n(A ∪ B ∪ C) and from (ii) to (i),
we get
Required number
= 243 – 93
= 150

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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board
Chapter 6 Permutations and Combinations
Miscellaneous Exercise 6 | Q 13 | Page 93