We have 9 parking spots, and 9 cars which will be parked randomly in these spots. 3 of the cars are sport cars, 3 of them trucks and 3 of them small cars. What is the probability that the 3 sport cars are parked next to each other?

So I already solved this using permutations once, and then again using combinations, but now I want to solve it using the multinomial coefficient. Let's call the event that the 3 sport cars are parked next to each other $A$. Then $$P(A) = \dfrac{n_A}{N}$$

$N= \dbinom{9}{3,3,3}$, as we're using multinomial coefficients.

Then $n_A = 7 \times \dbinom{6}{3,3}$ apparently, but I don't understand the $\dbinom{6}{3,3}$ term. Why does it matter how the remaining (non-sport) cars are ordered? I thought the point of combinations (binomial or multinomial) was that the order doesn't matter?


Let we number the parking slots with $1,2,\ldots,8,9$. If the three sport cars are next to each other, they occupy the slots $(1,2,3)$, $(2,3,4),\ldots,(7,8,9)$ (so $7$ possibilities) and they can be ordered in $3!=6$ ways. There are $6$ remaining slots: we may choose in $\binom{6}{3}=20$ ways the three slots occupied by trucks, then we have $3!$ orderings for the trucks and $3!$ orderings for the small cars.

Let we count, now, the total number of arrangements (unrestricted). We have $\binom{9}{3}=84$ ways for choosing the positions of the sport cars, then $\binom{6}{3}=20$ ways for choosing the positions of the small cars, and $3!\cdot 3!\cdot 3!$ ways for ordering trucks, small cars and sport cars in their corresponding parking slots. It follows that the wanted probability is given by

$$ \frac{7\cdot 20\cdot 3!\cdot 3!\cdot 3!}{84\cdot 20\cdot 3!\cdot 3!\cdot 3!}=\frac{7}{84}=\frac{7}{\binom{9}{3}}=\color{red}{\frac{1}{12}}\approx 8.33\%. $$

  • $\begingroup$ @YakSalTafri: $\binom{9}{3,3,3}$ is just the number of different anagrams of $TTTSSSsss$, i.e. $\frac{9!}{3!\cdot 3!\cdot 3!}$. $\endgroup$ – Jack D'Aurizio Sep 13 '16 at 21:14
  • $\begingroup$ I meant the $6 \choose 3,3$ term. Why is that there? The answer for $n_a$ implies that we don't care about the order of the sport car, however we do care about the order of the non-sport cars. Why? $\endgroup$ – YakSal Tafri Sep 13 '16 at 21:30
  • $\begingroup$ but then $n_A = 840$, so $\dfrac{n_A}{N} = \dfrac{1}{2}$ which is incorrect by a factor of $3!=6$ $\endgroup$ – YakSal Tafri Sep 13 '16 at 21:37
  • $\begingroup$ You have to decide first if you are accounting for the order of the same-type cars or not. Once you take an option, you have to get the same probability, $\frac{1}{12}$, with both approaches. $\frac{1}{2}$ is obviously wrong, it is an unreasonably high probability. $\endgroup$ – Jack D'Aurizio Sep 13 '16 at 21:42
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    $\begingroup$ Ah, I understand it now. The order thing confused me, I messed up. Thanks a lot. $\endgroup$ – YakSal Tafri Sep 13 '16 at 21:54

"As above, so bellow."   When comparing counts of atoms in the favoured space (the numerator) and the total space (denominator), they must be counted in the same general manner. †   Don't compare apples and orang-utans.

[ † PS: All atomic outcomes must have equal probability for this to be a valid method for measuring probability of the event. ]

By no coincidence, these three methods measure the same probability:

There are $9!$ ways to arrange nine singletons in list.   There are $7!3!$ ways to arrange six singletons and one triple in a list when the triple must remain connected. $$\dfrac{7!~3!}{9!}$$

There are $\binom 9 3$ ways to pick spots for the sports cars.   There are $7$ ways where these cars remain adjacent. $$\dfrac{~7~}{{9!}/{6!~3!}}$$

There are $\binom{9}{3,3,3}$ ways to pick places for all the vehicles.   There are $7\binom{6}{3,3}$ ways to pick places for them such that the sports cars remain adjacent.



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