A group of people waiting in an airport consist of 5 Americans, 4 Germans, 3 Canadians, 3 Italians, and 6 other nationalities. In how many ways can this group line up to board their aircraft if all the Americans are separated.
GOOD LUCK!
A group of people waiting in an airport consist of 5 Americans, 4 Germans, 3 Canadians, 3 Italians, and 6 other nationalities. In how many ways can this group line up to board their aircraft if all the Americans are separated.
GOOD LUCK!
I can't tell if this is a trick question, in the sense that: people are generally considered as distinguishable, regardless of nationality, so I think a normal interpretation for this question will be equivalent to having 5 Americans and 16 non-Americans. So the many nationalities are irrelevant and just there to throw us off. But this seems too sneaky.. So are we to treat Americans like red balls, Germans like green balls, etc.? But then it's not clear in another point: Are we to treat the 6 people of "other nationalities" as all distinct from each other, or all identical to each other? So, I think it's ambiguous which of these three interpretations the author intends.
Ah, the first interpretation then.
Yes, I suppose we should treat the question as written, and not worry about whether the author meant something different. I've gotten paranoid from spending time on sites where answers are judged by a machine, so it's necessary to understand the author's intent in order to get accepted by the judge. Lately I've come across a number of poorly written combinatorics problems.
C(17,5) ways to place Americans and non-Americans ignoring order, and 16! * 5! ways to permute.
And hello, Plato, long time no see.
Actually, answers for the second and third interpretations I mentioned are easily obtained from Plato's answer.
$\displaystyle \dfrac{\binom{17}{5}\cdot 16!}{4!\cdot 3!^2}$
and
$\displaystyle \dfrac{\binom{17}{5}\cdot 16!}{4!\cdot 3!^2\cdot 6!}$
Does that mean you solved it a different way? Note that you shouldn't post problems in this subforum if you don't know how to solve them yourself.
For the binomial coefficient, you can see it this way: _ X _ X _ X _ ... _ X _ where there are 16 X's that represent non-Americans, and 17 spaces where we can put the 5 Americans. If we remove the restriction that the Americans are separated it becomes $\displaystyle \binom{21}{5}$. (But with the restriction removed, then we can write it more simply without any binomial coefficients -- it's just to show how to transform the solution for one problem into a solution for another problem with smallest edit distance.)