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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 ?

  • 0
    Dice idemtical or all different? – 2017-01-21
  • 0
    Why are you using this method? Its tough and chances of mistake. – 2017-01-21
  • 0
    they are identical – 2017-01-21
  • 0
    that is what mind told me to do :D – 2017-01-21

1 Answers 1

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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.