# 7 cars are parked side-by-side. 4 of them are black. How many different ways are there of arranging the cars so that the 4 black cars are parked next to each other?

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We have 7 cars parked side-by-side and 4 of them are black. To work out the number of ways the black cars can be parked next to each other, we proceed as follows:

First, we need to use a formula for choosing r items in a row out of n (r<=n).

This is n - r + 1

Because there is the first way, eg in the case of 4

b b b b . . .

and then n - r more ways as we shift to the right by 1 place each time

So, we have n = 7 spaces and r = 4 black cars. The number of ways of getting 4 black cars in a row is

7 - 4 +1 = 4 ways

Secondly, we need to think of the number of ways the individual 4 black cars can be arranged amongst themselves given that 4 black cars are arranged in 4 particular spaces.

This is the number of *permutations* of the 4 black cars. This is equal to

4! = 4 x 3 x 2 x 1 = 24

There are 4 ways of placing the first car and 3 ways of placing the second, 2 ways of placing the third and then only 1 way of placing the fourth.

For each possible arrangement of 4 black cars in a row there are 24 ways of arranging the individual cars amongst themselves, so in total there are

24 * (n - r + 1) = 24 * (7 - 4 + 1) = 24 * 4 = 96 ways

**Answer: there are 96 ways**