A fair dice is thrown six times and the list of numbers showing up is noted. The probability that among the numbers 1 to 6 only 4 nu... Question : 
A fair dice is thrown six times and the list of numbers showing up is noted. Now how to find the probability that among the numbers 1 to 6 only 4 numbers appear in the list 
Please suggest how to solve such problems on probability this will be of great help to me. 
Thanks.
 A: There are 2 forms an answer can take:
Form #1:
ABCDAA (1 triplet (AAA) and 3 singles(BCD))
Form #2:
ABCDAB (2 doubles (AA and BB) and 2 singles (C and D))

For each form, we will multiply 3 things to count all the
  possibilities for that form:
  1. # of ways to choose 4 things from 6
  2. # of ways to select 2 things from 4 AND satisfy the conditions of the form
  3. # of ways those 6 things can be arranged

Calculation for Form #1
1. 6 Choose 4 = 15
2. We can choose AA, BB, CC, or DD, so 4
3. We can arrange 6 things in 6! ways, but we must divide to account for the tripled value, so we have 6!/3! = 120
15 * 4 * 120 = 7,200 ways for form #1
Calculation for Form #2
1. 6 Choose 4 = 15
2. 4 Choose 2 = 6
We can choose AB, AC, AD, BC, BD, or CD, but not BA, CA, DA, CB, DB, and DC because we would be double counting. Remember, we will multiply in step 3 to account for all arrangements, so that will cover these cases.
3. We can arrange 6 things in 6! ways, but we must divide to account for the two doubled values, so we have 6!/(2!*2!) = 180
15 * 6 * 180 = 16,200 ways for form #2
So, 7,200 ways for form #1 + 16,200 ways for form #2 = 23,400 ways total to get exactly 4 distinct numbers with 6 rolls of a die.
There are 6^6 = 46,656 possible outcomes for rolling a die 6 times, so the probability for rolling exactly 4 distinct numbers in 6 rolls is 23,400/46,656 = 50.154%
