Carcass wrote:
Jane must select three different items for each dinner she will serve. The items are to be chosen from among five different vegetarian and four different meat selections. If at least one of the selections must be vegetarian, how many different dinners could Jane create?
A 30
B 40
C 60
D 70
E 80
there are many ways to solve the problem -((Atleast one veg means 1 Veg or 2 Veg or all 3 Veg dishes.))
SInce Jane always have 1 veg dish, we can take the following combination-
1veg dish AND 2 non-veg dishes = 5C1 * 4C2 = 5*6 = 30 ways
or 2veg dishes AND 1 non-veg dish = 5C2 * 4C1 = 10*4 = 40 ways
or 3 veg dishes = 5C3 = 10 ways
Total no. of ways Jane can create different dinners = 30+40+10 = 80 ways
Alternative way-
The number of ways to choose 3 items from the total of 9 items is 9C3 = 84.
The number of ways to choose only meat is 4C3 = 4.
Therefore, there are 84-4 = 80 different dinners
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