JEE Main 2026 — 6 April, Evening Shift — Question 28

Total number of ways to draw 6 pairs from 12 balls without replacement: $\frac{12!}{(2!)^6 \cdot 6!}$. Number of favorable ways: each pair must be one blue and one green. Arrange 6 blue and 6 green in a sequence, then pair consecutive balls: $6! \times 6!$. Probability = $\frac{6! \times 6!}{\frac{12!}{(2!)^6 \cdot 6!}} = \frac{6! \times 6! \times (2!)^6 \times 6!}{12!}$. Simplify: $\frac{(6!)^3 \times 2^6}{12!}$. Compute: $6! = 720$, $(6!)^3 = 720^3 = 373248000$, $2^6 = 64$, numerator = $373248000 \times 64 = 23887872000$. $12! = 479001600$. Probability = $\frac{23887872000}{479001600} = \frac{238878720}{4790016} = \frac{16}{231}$. Thus, the probability is $\frac{16}{231}$, which corresponds to option C.