“페퍼로니 피자 10판이랑 파인애플 피자 10판이요🍕🍕” 중복조합 개념 쉽게 이해하기 I 정승제의 확률과 통계

Combinations with Repetition

Combinations with repetition count ways to choose a set number of items from several types when the same type can be chosen more than once and the order of choices does not matter.

How They Differ from Ordinary Combinations

When choosing people, you cannot select the same person more than once. But when choosing food types, you can choose the same type repeatedly—for example, ordering several pizzas of one flavor.

Because order does not matter, choosing pepperoni and cheese is the same selection as choosing cheese and pepperoni.

Food Examples

If you choose two scoops from three ice-cream flavors and repeats are allowed, there are six possible selections: two scoops of each flavor, plus one scoop of each possible pair of flavors.

Ordering a set number of pizzas from several flavors is also a combinations-with-repetition problem. The number of pizzas of each flavor can vary, but the total must equal the number ordered.

The “Sticks and Circles” Method

1. Represent each selected item with a circle. For (r) selections, draw (r) circles.
2. Represent the divisions between (n) available types with (n-1) sticks.
3. Arrange the circles and sticks in a row. Each arrangement shows how many items of each type were selected.
4. There are (r+n-1) positions altogether. Choose which (r) positions hold the circles—or, equivalently, which (n-1) hold the sticks.

Formula

The number of ways to choose (r) items from (n) types, with repetition allowed, is

[ \binom{n+r-1}{r}=\binom{n+r-1}{n-1}. ]

For two scoops from three flavors:

[ \binom{3+2-1}{2}=\binom{4}{2}=6. ]

Speaker

Jeong Seung-jae is the instructor. No other clearly identifiable speakers are featured.