Interesting problem I can't figure out...?

Question

Say a car dealership sells one type of car. The car has three options, of which a buyer can get any combination (none, one, two, or all three). The car also comes in four colors.

If the dealership sold 100,000 of these cars, what is the MAXIMUM number of any one type of car (meaning a distinct set of options/color) that you can GUARANTEE were sold.

At first I thought this was easy, then I thought about it and am unsure of how to solve it.

Answer 1

The key to doing a problem like this is breaking it down into manageable steps, and I'll try to help you with that.
The first step is to figure out how many different configurations are available for the car. (I am using the word configuration here to mean a distinct set of options/color)
We'll starts with the three options, and call them A, B, and C. There are eight different possibilities for these different options: none, A, B, C, AB, AC, BC, ABC.
There are also four colors, any of which can go with any set of options. The total number of configurations for the car is the number of different possibilities of options multiplies by the number of colors, which is 8 x 4 = 32.
The second step is to use this information to figure out the maximum number of cars you can guarantee were sold. It might be helpful to look at a couple of cases:
If you only sold 32 cars, it would be possible to sell one of each configuration, but if you sold 33 cars, you would need to sell at least two of one of the configurations. (The first 32 people could buy different configurations, but the 33rd would need to buy a configuration that matches one of the others.)
If you sold 64 cars, the maximum number of cars of any individual configuration sold would be two, because it is possible to sell exactly two configurations of each car. The 65th car sold, however, would have to match one of the configurations. Thus, for 65 cars sold the maximum number of cars guaranteed to fit a single configuration would be 3.
In the general case, for any multiple of 32 the number of cars sold could be found by (cars sold)/32. For any number that does not divide evenly by 32, we can take (cars sold)/32, and receive an answer in decimal form. However, we cannot sell a decimal of a car, so we will round up (regardless of what the decimal is).

So the key to solving this problem would be to figure out how many times 32 goes into 100,000.

Your answer would be 100,000/32 rounded up to the next whole number, which I don't know because I don't have a calculator.

P.S. If this is a homework problem, read the whole thing before looking at the answer or you won't know how to solve it for the test.

Answer 2

There are 2^3 (eight) ways to choose the 3 options (include option 1 or not, include option 2 or not, include option 3 or not). Then there are 4 choices of colors.

8 x 4 = 32 combinations. If there is nothing saying that each combination must exist, then each combination might have 0 people opting for that. So the maximum guaranteed would be 0.

Now if you are asking, what are the maximum number of cars that might be sold with the same combination, it would be 1/32 of 100000, if every combination is bought evenly. That's 3125 cars.

Answer 3

the guarantee part doesn't make sense with relation to the maximum number. But if 100,000 cars were sold, wouldn't the maximum for ANY distinct car sold be 100,000?

Answer 4

4 options 0X Y Z
4 colours A B C D

This means that there are 32 options

A0 AX AY AZ AXY AXZ AYZ AXYZ
B0 BX BY BZ BXY BXZ BYZ BXYZ
C0 CX CY CZ CXY CXZ CYZ CXYZ
D0 DX DY DZ DXY DXZ DYZ DXYZ

If you only sold 1 option of 31 of these cars, the 32nd option would have sold 99,969 cars

Answer 5

I don't know how you could guarantee that a specific set of options would be sold at all.
Zero?

Answer 6

i would say about 1,600,000