# Given constraints, in how many ways can actors be chosen for roles?

Given $$13$$ actors and $$6$$ unique roles, in how many ways can the actors be assigned a role if a certain actor (Alan) will not join if another actor (Betty) joins?

My method was to compute total unrestricted number of ways less the number of ways when they're both in it. The first term is $$\binom{13}{6} 6!$$ and the second term is $$\binom{11}{4} 6!$$. The numerical answer I get is $$997920$$ but the correct numerical answer given is $$1116720$$.

• 997920 looks right to me – Hagen von Eitzen Apr 19 at 8:16
• The result 1116720 would fit the task that we only prohibit Betty in a more prominent role than Alan – Hagen von Eitzen Apr 19 at 8:20
• What does being in a more prominent role mean? Also I realise that I get their answer only if I divide my second term by 2 (or 2! ?). – OneGapLater Apr 19 at 8:22
• The given answer will be true if a Alan has problem with Betty but not the other way around, assuming that they join one by one. – SinTan1729 Apr 19 at 9:12
• I think that's what they imply. However, I don't see why my answer doesn't accommodate for that. – OneGapLater Apr 19 at 9:27

Interpretation: Alan and Betty cannot both be cast.

You are correct under this interpretation. We can confirm your answer using a different approach. Observe that either exactly one of Alan or Betty is in the cast or neither Alan nor Betty is in the cast.

Exactly one of Alan or Betty is in the cast: Select whether Alan or Betty is in the cast. Select five of the other eleven actors to be in the cast. Assign roles to the six selected actors. $$\binom{2}{1}\binom{11}{5}6!$$

Neither Alan nor Betty is in the cast: Select six of the other eleven actors to be in the cast. Assign roles to the six selected actors. $$\binom{11}{6}6!$$

Total: Since the two cases above are mutually exclusive and exhaustive, the number of ways of assigning roles to the actors is $$\binom{2}{1}\binom{11}{5}6! + \binom{11}{6}6! = \binom{11}{5}6!\left[\binom{2}{1} + 1\right] = 3\binom{11}{5}6! = 997920$$ as you found.

Interpretation: Alan will not join if Betty has been cast first.

Under this interpretation, Alan and Betty can both be in the cast if Alan is cast first.

Then there are three possibilities:

1. Exactly one of Alan or Betty is in the cast.
2. Neither Alan nor Betty is in the cast.
3. Alan and Betty are both in the cast, because Alan is cast first.

We have covered the first two cases above.

Alan and Betty are both in the cast, because Alan is cast first: Ignoring the order of casting for the moment, if Alan and Betty are both selected, we must select four of the remaining eleven actors. We then assign roles to the six actors. This can be done in $$\binom{11}{4}6!$$ ways. However, by symmetry, in half of these assignments, Betty is cast before Alan. Thus, Alan will only join the cast in half of these assignments. Thus, there are $$\frac{1}{2}\binom{11}{4}6! = 118800$$ ways to assign the roles so that both Alan and Betty are cast.

Total: Since these three cases are mutually exclusive and exhaustive, the roles may be cast if Alan will not join if Betty has already been cast is $$\binom{2}{1}\binom{11}{5}6! + \binom{11}{6}6! + \frac{1}{2}\binom{11}{4}6! = 1116720$$ If this is the intended interpretation, the wording of the question could have been clearer.