A 9 dice are rolled at the same Time?what is the possible number of combinations, so that we have ( exactly )one 1 or at (least) a 4 or (both) A 9 dies are rolled at the same  Time !
what is the  possible  number of combinations, so that we have ( exactly )one 1  or at (least) a 4 or (both)
My solution ;
Let C = exactly one 1 
B = at least a 4 
$Let  A = C  \vee  B \vee ( C \wedge B) $
$ \neg A = \neg C \wedge \neg B \wedge (\neg B \vee \neg C ) $ 
$ \neg A = \neg C \wedge \neg B \wedge \neg C  \vee \neg C \wedge \neg B \wedge \neg B$
$ \neg A = \neg C \wedge \neg B $
$ \neg C$ = no 1 $\vee$ at least two 1's 
$ C_{1} $ = no 1 ; 
$ C_{2} $ = at least two 1's 
$ \neg C = C_{1} \vee  C_{2} $ 
$ \neg B$ = no  4 
$ \neg A =  (C_{1} \wedge \neg B ) \vee  (C_{2}  \wedge \neg B ) $
$  |C_{1} \wedge \neg B| $ = no 1 and no 4  = $ \binom{9+4-1}{9} $
$ |C_{2} \wedge \neg B| $ =  at least two 1's and no 4 = $ \binom{7+4-1}{
7}+\binom{6+4-1}{6}+\binom{5+4-1}{5}+\binom{4+4-1}{4}+\binom{3+4-1}{3}+\binom{2+4-1}{1}+\binom{1+4-1}{1}+1 $
$ C_{1} \wedge \neg B $  and $ C_{2} \wedge \neg B $ are disjoint set
$|\neg A| $ = $ |C_{2} \wedge \neg B| +|C_{1} \wedge \neg B| $
$ |A| = \binom{9+6-1}{9}- |\neg A|$
Is my answer correct ? 

Rolling 9 identical dice, in general, corresponds to placing 9 identical balls in 6 boxes.
To take care of the conditions placed, use stars and bars for the following $2$ cases:


*

*Zero $1's$, at least one $4$:
Consider as if only boxes $2-6$ left, with one $4$ preplaced $\to\binom{8+5-1}{8}$

*Exactly one $1$:
Preplace one $1$, again consider as if only boxes $2-6$ left $\to \binom{8+5-1}{8}$
Add up.
