A bag contains 6 blue and 6 green balls. Pair | vedclass

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(C) Total number of ways to partition $12$ balls into $6$ pairs is given by $\frac{\binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\bino

Vedclass · Accepted Answer

$\frac{16}{231}$

Vedclass · Answer

(C) Total number of ways to partition $12$ balls into $6$ pairs is given by $\frac{\binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}}{6!} = \frac{12!}{2^6 	imes 6!}$.Number of ways to form $6$ pairs such that each pair contains one blue and one green ball is $(6! 	imes 6!) = (6!)^2$, because we can pair the $6$ blue balls with the $6$ green balls in $6!$ ways, and the order of the $6$ pairs does not matter, but the calculation $(6!)^2$ accounts for the specific pairing arrangement.Probability $P = \frac{(6!)^2}{\frac{12!}{2^6}} = \frac{6! 	imes 6! 	imes 2^6}{12!}$.$P = \frac{720 	imes 720 	imes 64}{479001600} = \frac{518400 	imes 64}{479001600} = \frac{33177600}{479001600} = \frac{16}{231}$.

$A$ bag contains $6$ blue and $6$ green balls. Pairs of balls are drawn without replacement until the bag is empty. The probability that each drawn pair consists of one blue and one green ball is:

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