Question

12 normal dice are thrown once.The number of ways in which each of the values 2 , 3 , 4 , 5 and 6 occurs exactly twice is [1 ,1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , 6 can come in any order]

• A. $\frac{\left(12\right)!}{6}$
• B. $\frac{\left(12\right)!}{2^6.6!}$
• C. $\frac{\left(12\right)!}{2^6}$
• D. none

Answer 1

The correct option is C $\frac{\left(12\right)!}{2^6}$

Outcomes are 1,1,2,2,3,3,4,4,5,5,6,6

Number of ways =12!/(2!$\times 2!\times 2!\times 2!\times 2!\times 2!$)=$\frac{12!}{2^6}$