Browse Questions

The probability that in a throw of 12 dice 1,1, 2,2, 3,3, 4,4, 5,5, 6,6 will occur in any sequence is :

$\begin {array} {1 1} (1)\;\large\frac{12!}{2^66^{12}} & \quad (2)\;\large\frac{12!}{6^62^{12}} \\ (3)\;\large\frac{12!}{2^66^{6}} & \quad (4)\;None\: of \: these \end {array}$

Can you answer this question?

There are 6 faces on each of 12 dice. Hence the total number of
ways in which all dice can fall is $6^{12}$
Now the number of way in which 1,1, 2,2,....6,6 can be arranged on dice
= $\large\frac{12!}{(2!)^6}$
Hence the required probability is $\large\frac{12!}{2^66^{12}}$
Ans : (2)

answered Jan 2, 2014