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Also, in how many ways can we arrange them if $15$ people should be in one of the cars? I used ‘stars and bars’ method.

Let $x_i$ be number of people in car i;

So $x_1+x_2+x_3+x_4+x_5=80$

In the first question $x_1=15$, so $x_2+x_3+x_4+x_5=65$, that's $\frac{(65+3)!}{(65! \times 3!)}$

But I am stuck on question $2$. Because if we multiply $\frac{(65+3)!}{(65! \times 3!)}$ by $5$, we will get more arrangements, from what are possible because we count multiple times those arrangements when in at least $2$ cars there are exactly $15$ people.

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  • $\begingroup$ have you got to work out how many ways you can choose 15 from 80 to go in car 1 Then multiply that by the ways of spreading the remaining 65 over 4 cars? I can't see any element of double counting in that $\endgroup$ – Cato Feb 2 '17 at 16:49
  • $\begingroup$ But why multiply by 5. As its clear that in first car exactly 15 people. $\endgroup$ – Kanwaljit Singh Feb 2 '17 at 16:50
  • $\begingroup$ in the second one, exactly 15 in any car, there is a big danger of double counting - so the answer above is for 1) only $\endgroup$ – Cato Feb 2 '17 at 16:51
  • $\begingroup$ Also, the people are IDENTICAL, if we have 15 people in 1 car, it doesn't matter who they are., it matters only that they are 15. $\endgroup$ – Nagy Laszlo Feb 2 '17 at 16:53
  • $\begingroup$ are the people indistinguishable? As in we only care about how many people are in the cars? That doesn't make sense with (1) since it reduces the question to 4 cars and 65 people - or is that the point? If that is the point of the question, I agree with your answer to (1) $\endgroup$ – Cato Feb 2 '17 at 16:54
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With your assertion that only numbers matter for people, and cars are distinct, the first part is ok.

For the second part, apply stars and bars with inclusion-exclusion. (15 people in at least one car - 15 people in at least two cars + ....)

$=\dbinom51\dbinom{68}3 - \dbinom52\dbinom{53}2 + \dbinom53\dbinom{38}1 - \dbinom54\dbinom{23}0$

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  • $\begingroup$ Thank you.This seems right. I thought that with inclusion-exclusion, it would be too difficult. $\endgroup$ – Nagy Laszlo Feb 2 '17 at 17:15
  • $\begingroup$ You're welcome ! Btw, the question and its interpretation isn't as foolish as it appears. If, say, you are taking a group of people for a concert, who sits in which car is hardly a concern ! $\endgroup$ – true blue anil Feb 2 '17 at 17:34
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Of course the fact that the problem is about people rather than red balls or cases of beer suggests that you should consider that the people are distinguishable.

The question remains if the cars are distinguishable. I will assume that they are, and will also assume that the car that should have 15 people has already been chosen.

You start by picking 15 people for the first car: $$ {80\choose15}=\frac{80!}{15!(80-15)!}=6635869816740560. $$ Then you need the number of ways to fill exactly 4 cars with the remaining 65 people. Note that this is the number of ways to fill at most 4 cars minus the number of ways to fill at most 3 cars, i.e. $$ 4^{65}-3^{65}. $$ Therefore the number is $$ {80\choose15}\times(4^{65}-3^{65})=9032277882922331201019483936739005898712103023525609360 $$ or $\approx9\times10^{54}$.

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  • $\begingroup$ people are distinguishable, but then there is the question of simply considering how many people end up in each car - such as 'how many to a car will it be' - so there is also the indistinguishable case - it's a pity it can be so hard to see which would apply though - well that is hard in itself, but I mean which one is intended to apply. $\endgroup$ – Cato Feb 2 '17 at 17:07
  • $\begingroup$ In this particular problem, we consider people indistinguishable. $\endgroup$ – Nagy Laszlo Feb 2 '17 at 17:08
  • $\begingroup$ @Cato If the goal was to get you to consider that the objects are undistinguishable the problem would not be about people. It could be about cases of beer in 5 fridges :) $\endgroup$ – A.G. Feb 2 '17 at 17:09
  • $\begingroup$ @A.G. that is a good point, which I agree with $\endgroup$ – Cato Feb 2 '17 at 17:15
  • $\begingroup$ That made me think that you should ask the same question about cars. I edited my solution accordingly. If cars were indistinguishable the problem would probably be harder because more than one car could happen to carry 15 people... $\endgroup$ – A.G. Feb 2 '17 at 17:19

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