I had this confusion about permutation and combination. I got it clarified.

- permutation, if the order of objects is important. Means we consider all possible orders.
- combination, if the selection of objects is important. Means we consider this object with that object.

Ex:- I have a plate of 4 cups with different dishes (D1,D2,D3,D4), I want to taste only two dishes and see how the over all taste effect my impression on dishes.

- Combination,. (D1,D2), (D1,D3), (D1,D4), (D2,D3),(D2,D4),(D3,D4) => 6 combinations. 4C2 is the representation.
- Permutation, for each combination, I can taste in two ways. For example, (D1,D2) can be tasted in order (D1,D2) or (D2,D1). so total there are => 6 combination * 2 permutation for each combination = 12 permutations. 4P2 is the representation

Formulas:

NCr = N!/ ((N-r)! * r!)

NPr = N!/(N-r)!

** Which one is the permutation or combination problem?**

**1.** Doctor suggested to eat daily two verity of fruits and I have 4 apples, 3 oranges , 12 bananas and 6 chikku in my fridge. How many ways I can select two verity of fruits?

**Ans:** There are 6** combinations** possible for two verity of fruits from 4 verities. But there are multiple objects in each verity. So we need to consider all those objects when we select. For combination of 4 apples (A1, A2,A3,A4) and 3 oranges (O1,O2,O3), we can have combinations => (A1,O1), (A1,O2),(A1,O3) (A2,O1), (A2,O2),(A2,O3) (A3,O1), (A3,O2),(A3,O3)(A4,O1), (A4,O2),(A4,O3) (12 combinations). This is represented ad 4C1 * 3C1

Total number of ways: 4C1 * 3C1 + 4C1 * 12C1 + 4C1 * 6C1 + 3C1 * 12C1 + 3C1 * 6C1 + 12C1 * 6C1

If the problem says any two fruits ( Can be same fruit) then the number of ways will be => Add 4C2 + 3c2 + 12C2 + 6C2 to the above total.

**2.** IPL series has 8 teams and how many games will they play given two teams participate in a game?

**Ans:** This is a combination problem.

Choose 2 teams from 8 teams = 8C2 = 28 Games.

**2a. **Each team has their home ground, so total 8 grounds. If we put a constraint that each team has to play in other team’s ground.

**Ans: **For each combination, we have to play two games, For example, IND ground (IND vs SA), SA ground (SA vs IND). So total = 28 combinations * 2 = 56 Games. You can also treat this as **permutation problem**. 8P2 =56. Because the order of team comes in to picture.

**3.** I have 4 math books but my book shelf has space for only three books. How many ways I can arrange these books?

**Ans: **When we say arrange, the includes the order.

B1,B2,B3,B4 books in (B1,B2,B3), (B1,B2,B4), (B2,B3,B4) and (B1,B3,B4) => So 4C3 combinations => 4 combinations.

For combination (B1,B2,B3), we can arrange (B1,B2,B3),(B1,B3,B2),(B2,B1,B3),(B2,B3,B1),(B3,B2,B1)(B3,B1,B2) => orders. 3P3 = 6 orders

so total 4 * 6 = 24 arrangements.

**This is a permutation problem**, where we want to arrange 4 math books in 3 slots. Here we want all possible order of arrangements of books. The above explanation is represented as 4P3 = 24 arrangements.