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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=B
Total 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(2C12C12C12C1)=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(2C12C1)]=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(2C22C2)=240+15=255

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