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Question

Mathematics Question on permutations and combinations

Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five balls. In how many ways can we place the balls so that no box remains empty?

A

5050

B

100100

C

150150

D

200200

Answer

150150

Explanation

Solution

Let the boxes be marked as A,BA, B and CC. We have to ensure that no box remains empty and all five balls have to put in. There will be two possibilities : (i) Any two box containing one ball each and 33 rd box containing 33 balls. Number of ways =A(1)B(1)C(3)=5C14C13C3= A (1) \,B(1)\,C(3) =\, ^{5}C_{1} \cdot\, ^{4}C_{1} \cdot\,^{3}C_{3} =541= 5 \cdot 4 \cdot 1 =20 = 20 (ii) Any two box containing 22 balls each and third containing 11 ball, the number of ways =A(2)B(2)C(1)=5C23C21C1 = A(2)\, B(2)\, C(1) =\, ^{5}C_{2} \cdot\, ^{3}C_{2} \cdot\, ^{1}C_{1} =10×3×1= 10 \times 3 \times 1 =30= 30 Since, the box containing 11 ball could be any of the three boxes A,B,CA, B, C. Hence, the required number of ways =30×3=90= 30 \times 3 = 90. Hence, total number of ways =60+90= 60 + 90 =150 = 150.