There are 5 sedan Cars and 3 Hatchback cars all of different models. In how many ways they can be arranged in a row for an exhibition so that① no two hatchback cars are together

To solve this problem, we can use the concept of permutations.

First, let's consider the total number of ways to arrange all 8 cars (5 sedans and 3 hatchbacks) without any restrictions. The total number of arrangements is given by 8!, which is the factorial of 8.

Total arrangements without restrictions=8!

Now, let's consider the cases where two hatchback cars are together. Treat the two hatchback cars as a single unit (since their relative order among themselves doesn't matter), so we have 7 "units" in total (3 sedans, 1 hatchback unit, and 1 hatchback car).

The total number of arrangements in this case is 7!

because there are 7 units, and within each unit, there are different arrangements.

Now, within the hatchback unit, there are 2!

ways to arrange the two hatchback cars.

So, the total number of arrangements with two hatchback cars together is7!×2!

Finally, to find the number of arrangements where no two hatchback cars are together, subtract the above result from the total arrangements without restrictions:

Number of arrangements without two hatchback cars together8!−(7!×2!)

Number of arrangements without two hatchback cars together=8!−(7!×2!)

Now, you can calculate this value to find the answer.

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