1 . If a fair die is to be tossed **12 times** , the probability of getting **1,2,3,4,5**

**and 6 points** exactly **twice **each is

**(apologies cant find decent latex software)**

0.166 = **fraction 1 in 6**

**(12!** ** /** ** 2!2!2!2!2!2!** ) **(0.166*2 ) (0.166*2) (0.166*2) (0.166*2) (0.166*2) (0.166*2) = 0.00344**

** question 1**

does the **formula **used account for any order of the 2 of each ,meaning

there is no order

**3 3 , 4 4 , 6 6 , 5 5 ,1 1, 2 2** is the same as

**1 1, 2 2, 3 3, 4 4, 5 5 , 6 6** and so on , it accounts for all the possible combinations of the **2**

of each ?.........

if so ...

**question 2**

what would be the formula for the exact order

the first is **1** , second is **1 ** , the third is** 2** , the fourth is **2** etc in perfect running order?