How many ways can the cars be arranged if there are at least two cars of a particular brand in a row?

Question

I have the following exercise. Assume you have 5 cars of brand A, 6 cars of brand B and 5 cars of brand M. Each car is considered unique i.e no repetitions.

How many ways can I arrange the cars where there are at least two cars brand M in a row.

I started to evaluate the problem in the following way. As we have 5 cars brand M, and the restriction is at least 2, it means that the complement would be never two cars or more in a row. From there I calculated that scenario which was:

$$\frac{11!}{(6!*5!)} * \frac{12!}{(12-5)!*5!}$$

Then I subtracted the result from the total of options which would be 16!.

Is that correct??

Answer 1

Your strategy is correct, but you forgot to account for the fact that each car is unique.

There are $16!$ ways to arrange the cars. We must subtract those arrangements in which no two brand M cars are adjacent. We can line up the other eleven cards in $11!$ ways, creating twelve spaces in which to place the brand M cars, ten between successive cars and two at the ends of the row. $$\square c \square c \square c \square c \square c \square c \square c \square c \square c \square c \square c \square$$ To ensure that we separate the brand M cars, we must choose five of these twelve spaces in which to place a brand M car. The five brand M cars can be arranged in the selected spaces in $5!$ ways. Hence, the number of arrangements in which no two brand M cars are adjacent is $$11!\binom{12}{5}5!$$ Hence, the number of arrangements of the sixteen cars in which at least two of the brand M cars are adjacent is $$16! - 11!\binom{12}{5}5!$$ Had you multiplied your count of the arrangements in which no two brand M cards are adjacent by the $5!$ ways of arranging the brand A cars, $6!$ ways of arranging the brand B cars, and $5!$ ways of arranging the brand M cars, you would have obtained the correct answer. Observe that $$\frac{11!}{5!6!} \cdot \frac{12!}{5!7!} \cdot \color{red}{5!6!5!} = 11!\binom{12}{5}5!$$

Answer 2

You have the correct approach (enumerate the complement and subtract), but the wrong working. I will go through finding the number of arrangements with no two adjacent brand-M cars more carefully.

There are $6$ spaces between and around the $5$ cars of brand M: $$\square M\blacksquare M\blacksquare M\blacksquare M\blacksquare M\square$$ In the $4$ middle spaces, there must be at least one of the $11$ other cars; the end spaces do not have this restriction. Thus, ignoring the distinction between same-brand cars for the moment, we need to place $7$ cars in $6$ spaces with possibly $0$ cars in a space. By stars and bars the number of ways to do this is $$\binom{7+6-1}7=\binom{12}7$$ Given an arrangement of the brands, we can permute the individual cars in $5!11!$ ways. Thus there are $\binom{12}75!11!$ inadmissible arrangements where no two brand-M cars are adjacent.