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

The correct option is C

$\frac{12!}{{(2!)}^6{6}^{12}}$

Explanation for the correct option.

Step1.Calculate the possible outcomes:

Number of times die rolled $=12$

Number of faces on the die $=6$

Total number of ways each face appears$=6^{12}$

Step2. Calculate the number of favorable outcomes:

The total possibilities of each face turn twice $=\left\{1,1,2,2,3,3,4,4,5,5,6,6\right\}$

As per the given condition,

Favorable outcomes $=\frac{{}^{12}C_2}{2!}x\frac{{}^{10}C_2}{2!}x.........x\frac{{}^{2}C_2}{2!}$

$=\left(\frac{12!}{2!\times 10!\times 2!}\right)\left(\frac{10!}{2!\times 8!\times 2!}\right)..................\left(\frac{1}{2!}\right)$

$=\frac{12!}{{\left(2!\right)}^6}$

Step3. Calculate the required probability:

Required probability $=\frac{12!}{{\displaystyle \frac{{\left(2!\right)}^6}{6^{12}}}}\left[\mathbf{\because }\mathit{P}\mathit{r}\mathit{o}\mathit{b}\mathit{a}\mathit{b}\mathit{i}\mathit{l}\mathit{i}\mathit{t}\mathit{y}\mathbf{}\mathbf{=}\frac{\mathbf{N}\mathbf{u}\mathbf{m}\mathbf{b}\mathbf{e}\mathbf{r}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{f}\mathbf{a}\mathbf{v}\mathbf{o}\mathbf{u}\mathbf{r}\mathbf{a}\mathbf{b}\mathbf{l}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{u}\mathbf{t}\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{e}\mathbf{s}}{\mathbf{T}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{n}\mathbf{u}\mathbf{m}\mathbf{b}\mathbf{e}\mathbf{r}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{o}\mathbf{u}\mathbf{t}\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{e}\mathbf{s}}\right]$

$=\frac{12!}{{(2!)}^6{6}^{12}}$

Hence, Option(C) is the correct answer.