Question

A car dealer just took delivery on forty new cars. He plans to put four of these cars on display at the front of the lot. In how many ways can the dealer combine four of the forty cars if order IS important?

Answer 1

When order is important, we use permutations to find the number of ways to select and arrange the cars.

The number of permutations of selecting \( r \) items from \( n \) items is given by the formula:

\[
P(n, r) = \frac{n!}{(n-r)!}
\]

In this case, we have \( n = 40 \) cars and we want to select \( r = 4 \) cars. Plugging these values into the formula gives us:

\[
P(40, 4) = \frac{40!}{(40-4)!} = \frac{40!}{36!}
\]

This simplifies to:

\[
P(40, 4) = 40 \times 39 \times 38 \times 37
\]

Now, let's calculate this step by step:

1. Calculate \( 40 \times 39 \):
\[
40 \times 39 = 1560
\]

2. Multiply the result by \( 38 \):
\[
1560 \times 38 = 59280
\]

3. Finally, multiply by \( 37 \):
\[
59280 \times 37 = 2193360
\]

Thus, the total number of ways to combine and arrange four of the forty cars is

\[
\boxed{2193360}
\]