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 not important?

Answer 1

To determine the number of ways the car dealer can choose 4 cars from a total of 40 cars when the order of selection does not matter, we can use the combination formula. The formula for combinations is given by:

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

where \( n \) is the total number of items to choose from, \( r \) is the number of items to choose, and \( ! \) denotes factorial, which is the product of all positive integers up to that number.

In this scenario:
- \( n = 40 \) (the total number of cars)
- \( r = 4 \) (the number of cars to be chosen)

We substitute these values into the combination formula:

\[
\binom{40}{4} = \frac{40!}{4!(40-4)!} = \frac{40!}{4! \times 36!}
\]

Next, we can simplify \( \frac{40!}{36!} \):

\[
\frac{40!}{36!} = 40 \times 39 \times 38 \times 37
\]

Now, we substitute this back into our equation:

\[
\binom{40}{4} = \frac{40 \times 39 \times 38 \times 37}{4!}
\]

Next, we calculate \( 4! \):

\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]

Now we need to compute:

\[
\binom{40}{4} = \frac{40 \times 39 \times 38 \times 37}{24}
\]

Calculating the numerator:

\[
40 \times 39 = 1560
\]
\[
1560 \times 38 = 59280
\]
\[
59280 \times 37 = 2193360
\]

So we have:

\[
\binom{40}{4} = \frac{2193360}{24}
\]

Now perform the division:

\[
\frac{2193360}{24} = 91390
\]

Therefore, the number of ways the dealer can combine four of the forty cars is:

\[
\boxed{91390}
\]