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Question

# Suppose there are 6 pairs of different colored balls and 4 balls are selected. If same colored ball is different and A denote the number of ways when no pair of same colored ball is selected, B denote the number of ways exactly one pair of same colored ball is selected, C denote the number of ways at least one pair of same colored ball is selected, then

A
A=3C
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B
3B=2C
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C
CA=15
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D
A=B
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Solution

## The correct option is D A=BTotal number of balls =12 Case (1) If no pair of same colored ball is selected : We can select 4 pairs and then from each pair, either we can select first ball or second ball. Required ways, A=6C4⋅(2C1⋅2C1⋅2C1⋅2C1)=240 Case (2) If exactly one pair of same colored ball is selected : This can be done by first selecting any one pair of same colored ball out of 6 pairs. Then from rest 5 pairs, we have to select 2 balls with no pair. Required ways, B=6C1⋅[5C2⋅(2C1⋅2C1)]=240 Case (3) If at least one pair of same colored ball is selected : This means we can have exactly one pair or exactly two pairs, as 4 balls are selected. Required ways, C= number of ways of getting exactly one pair [Case (2)] + number of ways of getting 2 pairs =240+6C2⋅(2C2⋅2C2)=240+15=255

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