Question 1193917: Lucinda goes to the local pizzeria for dinner with a group of friends. The menu offers 4 types of soup, 5 main meals, 6 desserts and 3 different types of drinks. Lucinda will order 2 different courses and a drink. How many dinner options does she have?
Textbook answer is 222.
Can someone explain the solution of this problem?
I thought of
4*5*6 = 120 (for the first course)
3*4*5 = 60 (for the second course)
then i add them up to 180 and multiply it by 3 to get 540.
Found 3 solutions by greenestamps, ikleyn, math_tutor2020:
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
(1) The textbook answer you show doesn't make any sense. Are you looking at the answer for a different problem?
(2) The statement of the problem is deficient -- it doesn't define what a "course" is. Even if we assume that a "course" is a soup, a main meal, a dessert, and a drink, it is possible that the two different courses might contain, for example, the same soup.
Answer by ikleyn(52778)  (Show Source):
You can put this solution on YOUR website! .
In such problems, the answer is the product of some participating numbers.
But 222 = 2*111 = 2*3*37, and the number 37 is not among the input numbers.
Therefore, the answer in the textbook, mentioned in your post, is INCORRECT.
In our days, there are many pseudo-textbooks in the Internet, written/composed by pseudo-specialists for pseudo-students.
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
Define these three events
A = orders a soup
B = orders a main meal
C = orders a dessert
The notation AB means she orders a soup and a main meal.
For the case AB, there are 4 soups and 5 main meals giving 4*5 = 20 different outcomes here.
Imagine we had a table with 4 rows to represent the four different soups. Also, this table has 5 columns to represent the different main meals. There would be 4*5 = 20 inner cells to represent all possible soup & main meal combos. This is the reason why we multiply.
For case AC (soup+dessert), there are 4*6 = 24 outcomes
For case BC (MainMeal+dessert) there are 5*6 = 30 outcomes.
We have
AB+AC+BC = 20+24+30 = 74 ways to have Lucinda pick two different courses.
Something like AA, BB, or CC aren't allowed because she would order the same type of item (perhaps even two identical items) which isn't allowed.
She's only allowed 2 selections, and they must be different items. Admittedly the instructions are weirdly worded.
If she orders one drink (among 3 possible), and two different courses, then there are 3*74 = 222 different dinner options.
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