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A fair six-faced die is rolled 12 times. The probability that each face turns up twice is equal to
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Verified Answer
The correct answer is:
$\frac{12 !}{2^{6} 6^{12}}$
Required probability
$$
\begin{aligned}={ }^{12} C_{2} \times{ }^{10} C_{2} \times{ }^{8} C_{2} \times{ }^{6} C_{2} \times{ }^{4} C_{2} \times{ }^{2} C_{2} \times\left(\frac{1}{6}\right)^{12} \\=\frac{12 !}{10 ! \times 2 !} \times \frac{10 !}{8 ! \times 2 !} \times \frac{8 !}{6 ! \times 2 !} \times \frac{6 !}{4 ! \times 2 !} \\ \times \frac{4 !}{2 ! \times 2 !} \times \frac{2 !}{2 ! \times 1 !} \times\left(\frac{1}{6}\right)^{12} \end{aligned}
$$
$=\frac{12 !}{2^{6} \times 6^{12}}$
$$
\begin{aligned}={ }^{12} C_{2} \times{ }^{10} C_{2} \times{ }^{8} C_{2} \times{ }^{6} C_{2} \times{ }^{4} C_{2} \times{ }^{2} C_{2} \times\left(\frac{1}{6}\right)^{12} \\=\frac{12 !}{10 ! \times 2 !} \times \frac{10 !}{8 ! \times 2 !} \times \frac{8 !}{6 ! \times 2 !} \times \frac{6 !}{4 ! \times 2 !} \\ \times \frac{4 !}{2 ! \times 2 !} \times \frac{2 !}{2 ! \times 1 !} \times\left(\frac{1}{6}\right)^{12} \end{aligned}
$$
$=\frac{12 !}{2^{6} \times 6^{12}}$
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