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# Counting Partitions: Combinations - Finite Math

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• Thank you, Justin. It was helpful. but I reached multiple pools and scenarios I lost it. dime and quarters quiz I didn't solve it right any help?

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• ### Counting Partitions: Combinations - Finite Math

1. 1. 2.3Counting Partitions: Combinations
2. 2. Combination Combination is just like permutation – you are counting the number of ways to pick from a set without repetition of elements. The difference is that, for combination, order does not matter. Slot method can only be used when order matters; therefore, you cannot use slots for a combination problem.
3. 3. Formula n – number of things you are choosing from r – number of things you are choosing Since slots cannot be used, this formula is your only tool in solving combination problems
4. 4.  Let’s do an example combination problem. You order Mother Bear’s Pizza with a friend late at night. There is a special on a 3-topping pizza so you decide to go with that. There are 8 toppings to choose from. How many different pizzas can possibly be made? Simply apply the formula here. n = 8 (8 toppings to choose from) r = 3 (3 toppings being chosen to put on the pizza): Which comes to 56.
5. 5. Quiz 2.3 #1 How many ways can a committee of 3 be formed from 10 club members? A. 120 B. 240 C. 720
6. 6. Quiz 2.3 #1 How many ways can a committee of 3 be formed from 10 club members? A. 120 B. 240 C. 720 Answer: A
7. 7. Distinguishing betweenPermutation and Combination Permutation – order matters Combination – order doesn’t matter Let’s think of it another way – if the “slots” are distinguishable between each other, then order matters (P). In other words, if you rearrange the same elements, it becomes a different set.  Ex: A 3-digit number from {1, 2, 3, 4, 5}. This is a permutation, since digit 1 and digit 2 are “distinguishable.” If you switch them, it becomes a different number, i.e. 123 is different than 213.
8. 8.  On the other hand, if the slots are “indistinguishable,” then order doesn’t matter (C). In other words, if you rearrange the same elements, it is still the same thing. Ex: 3 toppings from 10 on a pizza. You cannot distinguish between topping 1 and topping 2; Pepperoni, Sausage and Ham is the same thing as Ham, Pepperoni and Sausage. One good way to identify between permutation and combination is to go ahead and draw and label the slots, and see if the slots are “distinguishable,” and if switching two elements from two slots affects the outcome. Remember, neither (P) or (C) can have repetitions. If there are repetitions, use neither of these two
9. 9.  Quiz 2.3 #2: Identify whether each problem is permutation, combination, or neither:
10. 10.  Quiz 2.3 #2: Identify whether each problem is permutation, combination, or neither: 1. Number of ways to form a committee of president, VP, and treasurer from 10 students.
11. 11.  Quiz 2.3 #2: Identify whether each problem is permutation, combination, or neither: 1. Number of ways to form a committee of president, VP, and treasurer from 10 students. Permutation
12. 12.  Quiz 2.3 #2: Identify whether each problem is permutation, combination, or neither: 1. Number of ways to form a committee of president, VP, and treasurer from 10 students. Permutation 2. Number of ways to select 5 distinct roles for a play out of 10 potential actors
13. 13.  Quiz 2.3 #2: Identify whether each problem is permutation, combination, or neither: 1. Number of ways to form a committee of president, VP, and treasurer from 10 students. Permutation 2. Number of ways to select 5 distinct roles for a play out of 10 potential actors Permutation
14. 14.  Quiz 2.3 #2: Identify whether each problem is permutation, combination, or neither: 1. Number of ways to form a committee of president, VP, and treasurer from 10 students. Permutation 2. Number of ways to select 5 distinct roles for a play out of 10 potential actors Permutation 3. Number of ways to pick a hand of 5 cards from a deck of cards
15. 15.  Quiz 2.3 #2: Identify whether each problem is permutation, combination, or neither: 1. Number of ways to form a committee of president, VP, and treasurer from 10 students. Permutation 2. Number of ways to select 5 distinct roles for a play out of 10 potential actors Permutation 3. Number of ways to pick a hand of 5 cards from a deck of cards Combination
16. 16. Choosingfrom Multiple Pools There are 4 Democrats and 3 Republicans forming a committee with 2 Democrats and 2 Republicans. How many different committees can be formed? We have to treat the Democrats and Republicans separately. This is a combination problem. So we have C(4,2) for Democrats, and C(3,2) for Republicans. You multiply the two: C(4,2) x C(3,2) = 6 x 3 = 18 2 Dem out of 4 2 Rep out of 3
17. 17. Choosingwith Multi-Scenarios A group of 6 friends are thinking about a Spring break trip to Florida. At least 4 of them have to go in order to get the group discount ﬂight. How many groups can be formed such that they can get the discount? This is also a combination problem. For this problem, at least 4 of them have to go, meaning the qualifying events are when 4 or 5 or 6 of them can go. You have to treat each scenario separately. So we have C(6,4) if 4 of them go, C(6,5) if ﬁve go, and C(6,6) if six go. This time, you add them: C(6,4) + C(6,5) + C(6,6) = 15 + 6 + 1 = 22 Group of 4 Group of 5 Group of 6
18. 18. Golden Rule of Finite Here it is: AND you multiply, OR you add. Let’s take the previous two examples. When we need 2 Democrats AND 2 Republicans, we multiplied C(4,2) x C(3,2). When we need a group of 4 OR 5 OR 6, we added C(6,4) + C(6,5) + C(6,6).
19. 19. Quiz 2.3 #3 I have 5 dimes and 2 quarters in my pocket. I reach in and pick 3 coins at random. How many ways can the 3 coins be selected? A. 35 B. 70 C. 115
20. 20. Quiz 2.3 #3 I have 5 dimes and 2 quarters in my pocket. I reach in and pick 3 coins at random. How many ways can the 3 coins be selected? A. 35 B. 70 C. 115 Answer: A
21. 21. Multiple Pools AND Scenarios Brad and Angie have 6 children, 3 boys and 3 girls. Three of them have to do dishes tonight. How many ways can the three be selected so that there is at least one boy and at least one girl? First, we’ll draw out a table representing the spectrum of events, ranging from all from one pool to all from the other: 0boy 1boy 2boys 3boys 3girls 2girls 1girl 0girl Checks represents qualifying events, crosses represents non-qualifying events
22. 22. Multiple Pools AND Scenarios So, 2b1g and 2g1b are the qualifying events. Now, since either 2b1g or 2g1b will satisfy the condition, we add up the number of elements in those events: C(3,2)C(3,1) + C(3,1)C(3,2) = 9 + 9 = 18 2boys 1girl 1boy 2girls
23. 23. Complementary Selection Sometimes there are more checks than crosses. In that case, it will be quicker to take advantage of this equation: # elements qualify = Total # elements in SS - # elements not quality Lets consider the last example, but instead of at least one boy one girl, we need only at least one boy: 0boy 1boy 2boys 3boys 3girls 2girls 1girl 0girl
24. 24. Complementary Selection In this case, it’s easier if we take total # of elements in the sample space and subtract the # in non-qualifying events from it: C(6,3) – C(3,3)*C(3,0) = 20 – 1= 19 any 3 3girls AND 0boys
25. 25. Quiz 2.3 #4 I have 6 regular and 4 diet coke in my cooler. I pick 5 at random. What’s the number of ways I pick at least 1 diet? A. 66 B. 246 C. 846
26. 26. Quiz 2.3 #4 I have 6 regular and 4 diet coke in my cooler. I pick 5 at random. What’s the number of ways I pick at least 1 diet? A. 66 B. 246 C. 846 Answer: B
27. 27. Summary Deﬁnition:  Combination (order doesn’t matter, no repetition) How to ﬁnd the number of combinations  Formula How to distinguish between combination and permutation Problems for multiple pools and multi-scenarios  Golden rule of Finite
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