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. $A=3C$
• B. $3B=2C$
• C. $C-A=15$
• D. $A=B$

Answer 1

Answer: (C), (D)

$A=B$

Total number of balls $=12$

Case $\left(1\right)$ 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, $A={}^6{C}_4\cdot ({}^2{C}_1\cdot {}^2{C}_1\cdot {}^2{C}_1\cdot {}^2{C}_1)=240$

Case $\left(2\right)$ 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, $B={}^6{C}_1\cdot [{}^5{C}_2\cdot ({}^2{C}_1\cdot {}^2{C}_1)]=240$

Case $\left(3\right)$ 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 $\left(2\right)$] $+$ number of ways of getting $2$ pairs
$=240+{}^6{C}_2\cdot ({}^2{C}_2\cdot {}^2{C}_2)=240+15=255$