# Probability and Statistics, very basic stuff

• parwana

## Homework Statement

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?

## The Attempt at a Solution

1. 5x7x2= 70. Is this right?
2. 3x5= 15, Is this right?
3. 2x5= 10 for first group, 3x7=21 for second group. So total would be 10x21= 210 meals possible?
4. 4x5= 20, 4x7= 28, 1x2= 2. Then do u multiply all of these three values?

You don't appear to have the concept of "combination" down pat.

Look at just the first group, the one with five items. You correctly identified the number of ways to choose one item from this group of five (five ways, question 1), but you got all of the other combinations wrong. Now if there are five ways to choose one item, aren't there are five ways to choose all but one item (question 4).

How do you calculate the number of ways one can construct a subset of m elements from a set that contains n unique elements?

For example for problem 2, "2. How many meals are possible containing 3 foods from the first group and none from the other two groups?", (assuming that choosing the same food three times is not "3 foods"), there are 5 choices for the first food, then 4 for the second, and 3 for the third, a total of 5*4*3= 5!/(5-3)! choices.

For example for problem 2, "2. How many meals are possible containing 3 foods from the first group and none from the other two groups?", (assuming that choosing the same food three times is not "3 foods"), there are 5 choices for the first food, then 4 for the second, and 3 for the third, a total of 5*4*3= 5!/(5-3)! choices.

That's not correct. Halls is describing permutations, not combinations. For example, suppose the five items in the first group are labeled a to e. Thee meal comprising items c, b, and a is the same as the meal comprising items a, b, and c.

The correct expression for the number of subsets of size m chosen from a set comprising n distinct items is $n!/(m!\cdot(n-m)!)$.

Ouch! yes. My method treats "mashed potatoes, green beans, and corn" as a different choice from "green beans, mashed potatoes, and corn". Since there are 3!= 3*2*1= 6 ways to interchange the three chosen items you need to divide my previous answer by 3!= 6. Since I had given the answer as 5!/(5-3)!, the correct answer is 5!/((5-3)!(3!)), the binomial coefficient, sometimes called "5, choose 3".

Thanks for catching that D H!