Number of empty subway cars

Question

I have task:

There are $9$ passengers and they get into empty $5$-car subway train. What is the probability of the fact, that exactly two cars will be empty?

I know that the power of all possibilities is $5^9$. Then I thought to pick two empty cars that is ${5\choose 2}$ possibilities, then pick three people that get to the rest of the cars, that is $9\cdot 8\cdot 7$ and the rest of the people have $3^6$ possibilities. I checked it on my calculator and it seems that the power of this event is bigger than the power of $\Omega$. What is wrong?

Robert Z · Answer 1 · 2019-12-03T07:32:34.363

3

Yes, we have $\binom{5}{3}=\binom{5}{2}=10$ ways to choose the $3$ not-empty cars. Now you have to multiply this number by the number of surjective maps from $\{1,\dots,9\}$ to $\{1,2,3\}$, i.e. $S(9,3)=3^9-3\cdot 2^9+3=18150$ where $$S(a,b) = \sum_{k=1}^b (-1)^{b-k} \binom{b}{k} k^a.$$ Such number is strictly less than $9\cdot 8\cdot 7\cdot 3^6$ (you are overcounting the number of cases).

Hence the probability that exactly two cars will be empty is $$p=\frac{10S(9,3)}{5^9}=\frac{181500}{1953125}=0.092928.$$

edited Dec 03 '19 at 07:32

answered Dec 03 '19 at 07:18

Robert Z

147,345

1

Probably worth adding that your $18150 = 3! \times \left\lbrace{9\atop 3}\right\rbrace$ where $ \left\lbrace{,\atop ,}\right\rbrace$ represents a Stirling number of the second kind – Henry Dec 03 '19 at 07:39
Thank You very much for Your help! – mwrooo Dec 03 '19 at 07:52
@Henry Good point! Thanks. – Robert Z Dec 03 '19 at 10:50
@mwrooo I am happy that my answer was helpful. Note that the person who asked can mark one answer as "accepted". Please visit https://math.stackexchange.com/tour – Robert Z Dec 03 '19 at 11:02

score 1 · Answer 2 · answered Dec 03 '19 at 13:48

The three cars must have at least $1$ passenger each. So the cases can be easily calculated: $$\{1,1,7\} \Rightarrow {9\choose 1}{8\choose 1}3=216\\ \{1,2,6\} \Rightarrow {9\choose 1}{8\choose 2}6=1512\\ \{1,3,5\} \Rightarrow {9\choose 1}{8\choose 3}6=3024\\ \{1,4,4\} \Rightarrow {9\choose 1}{8\choose 4}3=1890\\ \{2,2,5\} \Rightarrow {9\choose 2}{7\choose 2}3=2268\\ \{2,3,4\} \Rightarrow {9\choose 2}{7\choose 3}6=7560\\ \{3,3,3\} \Rightarrow {9\choose 3}{6\choose 3}1=1680\\ $$ The sum is $18150$.

The three cars with passengers can be any of the five: ${5\choose 3}=10$.

Hence, there are $18150\cdot 10=181500$ ways to arrange $9$ passengers in $5$ cars so that exactly $2$ will remain empty. The probability is: $$P=\frac{181500}{5^9}\approx 0.09.$$

Number of empty subway cars

2 Answers2