Word Problem

[Jensi]

Hello,

A restaurant offers burritos on corn or a flour tortilla, 5 types of meats, 6 types of cheeses, and 3 different toppings. When ordering, customers can choose 1 type of tortilla, 1 meat, and 1 cheese. They can then add any of the 3 toppings. How many different burrito combinations are possible?

A. 180
B. 330
C. 480
D. 660
E. 1620

Answer is C. 480
2*(5)*(6)2*2*2 = 480
2 tortilla * 5 Meats * 6 Cheeses *2*2*2

What is the 2*2*2 at the end represent? Is it for topping?
How come its 2 for toppings?

Thank You

 

[MarkFL]

I would say:

[MATH]N=2\cdot5\cdot6\sum_{k=0}^{3}{3 \choose k}=2\cdot5\cdot6\cdot2^3=480[/MATH]
The \(2^3\) comes from the sum of the binomial coefficients. The customer can choose no toppings, 1 of the 3, 2 of the 3 or all of the 3, and all these sum to 8.

 

[Steven G]

You have 2 choices for the one type of tortilla
You have 5 choicses for the one type of meat
You have 6 choices for the one type of cheese
You have 7 choices for up to three topping (1=all three, 3= any two, 3=any one). I suspect that the author is allowing no topping as a choice
The answer is 2*5*6*7=420

 

[Steven G]

I would say:

[MATH]N=2\cdot5\cdot6\sum_{k=0}^{3}{3 \choose k}=2\cdot5\cdot6\cdot2^3=480[/MATH]
The \(2^3\) comes from the sum of the binomial coefficients. The customer can choose no toppings, 1 of the 3, 2 of the 3 or all of the 3, and all these sum to 8.

You must be correct, but I think that we should point out to the OP, that it makes no sense that you can choose no topping but you can't choose no meat or no cheese, etc

 

[pka]

I would say: [MATH]N=2\cdot5\cdot6\sum_{k=0}^{3}{3 \choose k}=2\cdot5\cdot6\cdot2^3=480[/MATH]The \(2^3\) comes from the sum of the binomial coefficients. The customer can choose no toppings, 1 of the 3, 2 of the 3 or all of the 3, and all these sum to 8.

While your are of course correct, if I were still in my former role as test reader/editor I would have vetoed this question or at least that distractor. Here I agree with Jomo that testing the multiplicative principle does need to be confused with binomial coefficients.

 

[Jensi]

I would say:

[MATH]N=2\cdot5\cdot6\sum_{k=0}^{3}{3 \choose k}=2\cdot5\cdot6\cdot2^3=480[/MATH]
The \(2^3\) comes from the sum of the binomial coefficients. The customer can choose no toppings, 1 of the 3, 2 of the 3 or all of the 3, and all these sum to 8.

Hello

How do i get the sum of 8?
I forget how to do the sum of the binomial coefficient.

Thanks

 

[MarkFL]

Hello

How do i get the sum of 8?
I forget how to do the sum of the binomial coefficient.

Thanks

There is 1 way to choose no topping, 3 ways to choose 1 topping, 3 ways to choose 2 toppings and 1 way to choose 3 toppings:

[MATH]1+3+3+1=8[/MATH]
I was referring to (the binomial theorem):

[MATH]\sum_{k=0}^{3}{3 \choose k}=\sum_{k=0}^{3}\left({3 \choose k}1^{3-k}\cdot1^{k}\right)=(1+1)^3=2^3=8[/MATH]

 

[Dr.Peterson]

The [MATH]2^3[/MATH] can also be obtained this way: there are 2 ways to choose whether to have topping 1; then 2 ways to choose whether to have topping 2; then 2 ways to choose whether to have topping 3. I think of it as the number of ways to check any (or none, or all) of three check boxes indicating which you want. Or, the number of subsets of {1, 2, 3}.