Combinations Question: Burgers

Question

A tray holds 3 cheese burgers, 5 bacon burgers and 6 egg burgers.

How many ways to choose at least one of each type of burger to a total of 4 burgers?

This is my method: ($3 \choose 1 $* $5 \choose 1 $* $6 \choose 1$) * $11\choose1$ = 990 I was thinking choosing 1 of cheese ($3\choose1$), one of bacon ($5\choose1$), one of egg ($6\choose1$) then $11\choose1$ for the last(random burger).

The solution was doing it case by case: ($3\choose1$ * $5\choose2$ * $6\choose1$) + ($3\choose2$ * $5\choose1$ * $6\choose1$) + ( $3\choose1$ * $5\choose1$ * $6\choose2$) = 495

Now I understand how the solution works, but I don't understand why my original method can't work though...

Note that the true result is exactly half of yours. You just counted each way twice. — DanielY, Jun 17 '14 at 11:06
@DanielY yes, I noticed that, however I don't know why I'm counting each way twice though... — user9856, Jun 17 '14 at 11:10

score 3 · Answer 1 · answered Jun 17 '14 at 11:09

3

Your method yields $\;495\cdot 2=990\;$ options because you didn't take into consideration repetitions: for example, and denoting the different burgers by $\;C,B,E\;$ , one of your options is $\;C_1,B_1,E_1, C_2\;$, which is the same as $\;C_2, B_1,E_1,C_1\;$ ...see?

You're counting every quartet of burgers twice, as any of the three kinds of burgers can be chosen out of the first three (the "at least one of each" part), and then as the fourth burger...

answered Jun 17 '14 at 11:09

DonAntonio

211,718
17
136
287

2

well...no doubt you've explained this better than me :) – DanielY Jun 17 '14 at 11:11
Yes, thank you :) Are combinations questions like these always have to be done case by case though? – user9856 Jun 17 '14 at 11:31

score 1 · Accepted Answer · answered Jun 17 '14 at 11:10

1

You counted each way to pick the burgers twice.

Let's say there are burgers of type x, y, and z (more confortable). We need to choose one burger from each type, and the fourth can be any type. But, take into consideration that the combination, for instance (x, y, z, x) is counted twice according to your solution, because taking one x burger first and then another x burger as the random burger is the same as doing it the other way around

Hope my blurry answer helps

answered Jun 17 '14 at 11:10

DanielY

979

Oh, I see now. Is there a simple rule to remember concerning this type of combinations questions? I'm not good at realising when I'm over-counting or when I'm not... Somehow the first method I think of (which is easy) is wrong. By the way, was it just a coincidence in this case that I got exactly double? Thank you! – user9856 Jun 17 '14 at 11:13
when the items you need to count you collection of them are the same (e.g. balls of the same color, burgers etc.) you must take into cosideration repititions. When the items differ from one another (numbered balls, for instance), or when the order matters, your approach is fine – DanielY Jun 17 '14 at 11:14

Combinations Question: Burgers

2 Answers2