Question

A fair six-faced die is rolled $12$ times. The probability that each face turns up twice is equal to

• A. $\frac{12!}{6!6!6^{12}}$
• B. $\frac{2^{12}}{2^6{6}^{12}}$
• C. $\frac{12!}{2^6{6}^{12}}$
• D. $\frac{12!}{6^2{6}^{12}}$

Answer 1

Answer: (C)

Required probability
$={}^{12}{C}_2\times {}^{10}{C}_2\times {}^8{C}_2\times {}^6{C}_2\times {}^4{C}_2\times {}^2{C}_2\times {\left(\frac{1}{6}\right)}^{12}$
$=\frac{12!}{10!\times 2!}\times \frac{10!}{8!\times 2!}\times \frac{8!}{6!\times 2!}\times \frac{6!}{4!\times 2!}\times \frac{4!}{2!\times 2!}\times \frac{2!}{2!\times 1!}\times {\left(\frac{1}{6}\right)}^{12}$
$=\frac{12!}{2^6\times 6^{12}}$