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

Firstly, total number all possible outcomes by rolling a die $12$ times = $6^{12}$ …..(i).

Secondly, when each face turns up twice, the outcomes are like $1,1,2,2,3,3,4,4,5,5,6,6$ though not necessarily in that order always. Now, we have to find out in how many ways these $12$ numbers can be arranged of which there $21s,22s,\dots \dots 26s$.{ Remember permutation of n objects of which r objects are of one type, s objects of another type, t objects are yet of another type is $\frac{n!}{(r!s!t!)}$

Hence, it can be arranged in $\frac{12!}{(2!{)}^6}$ ways ……(ii)

Therefore the required probability = $\frac{\left(i\right)}{\left(ii\right)}=\frac{12!}{2^6{6}^{12}}$