I need help with this probability question?
Suppose that a dietitian has 3 groups of foods, having 5 items in the first group, 7 items in the second group and 2 items in the third group.
1. How many meals are possible containing one food from each food group?
2. How many meals are possible containing 3 foods from the first group and none from the other two groups
3. How many meals are possible containing 2 foods from the first group, 3 from the second group and none from the third group?
4. How many meals are possible containing 4 foods from first group, 4 from the second and 1 from the third group?
Favorite Answer
2. 5*4*3 = 60
3. 5*4 * 7*6*5 = 4200
4. 5*4*3*2 * 7*6*5*4 * 2 = 201,600
To explain,
In the first problem, there are 5 choices for what to take from the first group. Then, no matter which was chosen there are 7 choices to decide among for the second, and then 2 choices for the last group. Multiply and you get the total number of possibilities.
In the second problem, choose any of the five items. Then choose one of the four that remain, then choose any of the three that are left. Multiply again.
The other two are simply combinations of these facts.
I hope this helps!
2. 1 meal
3. 2 meals
4. 1 meal
I am not sure of the answers..
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