Asked by Anonymous
A hamburger restaurant offers 2 types of buns, 3 types of meat burgers and 4 types of different toppings. How many different hamburgers can be prepared if the chef must choose 1 bun, 1 meat burger but can have no topping, one topping or more than one toppings?
Please verify if my answer is correct:
2C1x3C1x(4C0+4C1)/C9
Thanks in advance
Please verify if my answer is correct:
2C1x3C1x(4C0+4C1)/C9
Thanks in advance
Answers
Answered by
MathMate
Contrary to probability calculations, we do not need to divide by XC9 when finding the number of arrangements.
C(2,1)*C(3,1) is correct for the first two items.
For the toppings, since we are allowed 0 to 4 toppings, the number of choice available equals the power set, or 2^4.
The total number of arrangements is therefore
C(2,1)*C(3,1)*2^4.
C(2,1)*C(3,1) is correct for the first two items.
For the toppings, since we are allowed 0 to 4 toppings, the number of choice available equals the power set, or 2^4.
The total number of arrangements is therefore
C(2,1)*C(3,1)*2^4.
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