4 identical dice rolled at the same time what is the number of the possibility to have exactly two 1's ?
somehow i don t find the correct answer between these :
(1):
since it is " at the same time " then the order plays no role here then we have
1122
1123
1124
1125
1126
1133
1134
1135
1136
1144
1145
1146
1155
1156
1166
==> there number of possibility is 15
(2)
since it is " at the same time " then the order plays no role here then we have 11xx and x in [2,6] Then the number of the possibility would be $\binom{2+5-1}{2} = \binom{6}{2} =30 \neq 15 $ !! what was worng !
Your only mistake is that $\binom{6}{2}$ does not equal to $30$. The formula you used is for permutations: $$^{r+n-1}P_r=\frac{(r+n-1)!}{(r+n-1-r)!}=\frac{(r+n-1)!}{(n-1)!}=\frac{6!}{4!}=30$$ But what you should be looking for is: $$^{r+n-1}C_r=\binom{r+n-1}{r}=\frac{(r+n-1)!}{(r+n-1-r)!r!}=\frac{(r+n-1)!}{(n-1)!r!}=\frac{6!}{4!\cdot 2!}=15$$