Question

A fair six-faced die is rolled 12 times. Find the probability that each face turns up twice.
A.$\dfrac{{12!}}{{6!6!{6^{12}}}}$
B.$\dfrac{{{2^{12}}}}{{{2^6}{6^{12}}}}$
C.$\dfrac{{12!}}{{{2^6}{6^{12}}}}$
D. $\dfrac{{12!}}{{{6^2}{6^{12}}}}$

Answer 1

Hint: First find the total outcome of a fair die rolled 12 times, then find the outcome of each face turns off twice. Then find the permutation of these 12(outcome of each face turns off twice) objects and divide $\dfrac{{12!}}{{{2^6}}}$ by ${6^{12}}$ to obtain the required result.

Formula Used:
The outcome of a fair die roll n times is ${6^n}$ .
The permutation of n objects where p is of one type, q is of one type and r is of another type is $\dfrac{{n!}}{{p!q!r!}}$ .

Complete step by step solution:
The total outcome of a fair die rolled 12 times is ${6^{12}}$.
The outcome of each face turning twice is $\left\{ {1,1,2,2,3,3,4,4,5,5,6,6} \right\}$ .
The permutation of these 12 objects where 1 to 6 occur twice is
$\dfrac{{12!}}{{(2!).(2!).(2!).(2!).(2!).(2!)}}$
$ = \dfrac{{12!}}{{2.2.2.2.2.2}}$
$ = \dfrac{{12!}}{{{2^6}}}$
Divide $\dfrac{{12!}}{{{2^6}}}$ by ${6^{12}}$ to obtain the required result.
$ = \dfrac{{\dfrac{{12!}}{{{2^6}}}}}{{{6^{12}}}}$
$ = \dfrac{{12!}}{{{2^6}{6^{12}}}}$

Option ‘C’ is correct

Note: In this problem we have used the permutation of n objects where 3 of them of 3 types, students some time skip this step and as the number of outcomes of twice the faces are 12 they simply write the answer as $\dfrac{{12}}{{{6^{12}}}}$.