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Question

Let m denote the number of ways in which four different balls of green colour and four different balls of red colour can be distributed equally among 4 persons if each person has balls of the same colour and n be the corresponding figure when all the four persons have balls of different colour. Then the value of m+n is

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Solution

4 green balls (G1,G2,G3,G4) and 4 red balls (R1,R2,R3,R4) are to be distributed among 4 persons.

4 green balls can be divided into two equal
groups in 4!2! 2! 2!=4!8 ways
Similarly, 4 red balls in 4!2! 2! 2!=4!8 ways.
Number of four equal groups =(4!8)(4!8)
Now, we have 4 equal groups. Each equal group has 2 balls of the same colour. They can be distributed in 4 persons in 4! ways.
m=(4!8)(4!8)(4!)=216

Now, n=(4C1)2(3C1)2(2C1)2(1C1)2=(24)2
=576
[(4C1)21 out of 4 green and 1 out of 4 red can be taken in (4C1)(4C1)=(4C1)2
ways.]

Hence, m+n=216+576=792

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