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?
45,695
2,193,360
91,390
1,096,680
Answers
GPT-4o mini
Since the order is important for displaying the cars, we need to calculate the number of permutations of 4 cars selected from a total of 40 cars.
The formula for permutations is given by:
\[
P(n, r) = \frac{n!}{(n - r)!}
\]
where \( n \) is the total number of items to choose from, \( r \) is the number of items to choose, and \( ! \) denotes factorial.
In this case, \( n = 40 \) and \( r = 4 \):
\[
P(40, 4) = \frac{40!}{(40 - 4)!} = \frac{40!}{36!}
\]
This simplifies to:
\[
P(40, 4) = 40 \times 39 \times 38 \times 37
\]
Now we calculate that step by step:
1. \( 40 \times 39 = 1560 \)
2. \( 1560 \times 38 = 59280 \)
3. \( 59280 \times 37 = 2193360 \)
Thus, the total number of ways to combine four of the forty cars when the order is important is:
\[
\boxed{2193360}
\]
The formula for permutations is given by:
\[
P(n, r) = \frac{n!}{(n - r)!}
\]
where \( n \) is the total number of items to choose from, \( r \) is the number of items to choose, and \( ! \) denotes factorial.
In this case, \( n = 40 \) and \( r = 4 \):
\[
P(40, 4) = \frac{40!}{(40 - 4)!} = \frac{40!}{36!}
\]
This simplifies to:
\[
P(40, 4) = 40 \times 39 \times 38 \times 37
\]
Now we calculate that step by step:
1. \( 40 \times 39 = 1560 \)
2. \( 1560 \times 38 = 59280 \)
3. \( 59280 \times 37 = 2193360 \)
Thus, the total number of ways to combine four of the forty cars when the order is important is:
\[
\boxed{2193360}
\]