A car dealer has 7 new sedans she wants to display but room for only 5 on the showroom floor.

How many ways can she display the 7 cars on the showroom floor?

A.
5040

B.
2520

C.
35

D.
21

thanks

Question

A car dealer has 7 new sedans she wants to display but room for only 5 on the showroom floor.
 
How many ways can she display the 7 cars on the showroom floor?
 
 A.
5040
 
 B.
2520
 
 C.
35
 
 D.
21

thanks

mathmale · Answer

This sounds like:  "A car dealer has 7 new sedans she wants to display but room for only 5 on the showroom floor.  How many combinations from 7 cars, 5 taken at a time, are there?

We can write this mathematically as $$_{7}C _{5}$$

mathmale · Answer

How would you evaluate that?  If you're not sure, look up "combinatorics" or "combinations and permutations" on the 'Net or in your online learning materials.

jaydenthequiet · Answer

8?

mathmale · Answer

How did you get 8?  Are you familiar with the formula for calculating the number of combinations from n objects taken x at a time?

jaydenthequiet · Answer

I thought I was

mathmale · Answer

$$_{n}C _{x}=\frac{ n! }{ x!(n-x)! }$$

mathmale · Answer

Here, n = 7 and x = 5.

jaydenthequiet · Answer

$$\frac{ 7! }{ 5!(7-5)! }$$

mathmale · Answer

Good.  Note that 7! = 7*6*5! and that this reduces to 7*6 if you divide it by that 5! in the denominator.  That leaves 7*6 = 42 in the numerator; you still have to divide that by (7-5)!.  What's that?

mathmale · Answer

In other words, what is the value of$$\frac{ 42 }{ 2! }?$$

jaydenthequiet · Answer

21

mathmale · Answer

that looks reasonable to me.  What do YOU think?

jaydenthequiet · Answer

I just did it on my calculator and it says that twenty-one is correct

mathmale · Answer

Cool.  Congrats!

Have to get off the 'Net now.  Hope to work with you again soon.