Combinaciones | Ejemplo 2

Introduction to Combinatorial Concepts

Overview of the Course

• The course focuses on combinatorial concepts, starting with an example of combinations.

• This video presents a more challenging exercise compared to previous ones, encouraging viewers to review earlier videos for foundational understanding.

Identifying Combinations vs. Permutations

• It is essential to determine whether a problem involves permutations or combinations; in this case, it is confirmed as a combination.

• A key indicator that a problem is a combination is that the order of selection does not matter.

Understanding the Problem Statement

Problem Description

• The task involves forming a committee from 12 men and 10 women, specifically selecting 5 men and 5 women.

• The distinction between groups (men and women) must be noted since selections are made separately for each group.

Importance of Order in Selection

• The speaker emphasizes that the order of selection does not affect the outcome; all selected individuals will perform the same role within the committee.

• An example illustrates that rearranging selected individuals does not create different committees, reinforcing that order is irrelevant in this context.

Calculating Combinations

Setting Up for Calculation

• To calculate combinations, one must identify 'n' (total number of items) and 'r' (number of items to choose). For men: n = 12 and r = 5.

Combinatorial Selection of Men and Women

Introduction to Combinations

• The discussion begins with the selection of 5 women from a group, indicating that combinations will be calculated for both men and women.

• Emphasizes the multiplication rule in combinatorics, stating that results from selecting men and women will be multiplied together.

Calculating Combinations for Men

• The formula for combinations is introduced: C(n, r) = n!/r!(n-r)! . Here, n is 12 (men), and r is 5.

• Simplification process explained by expressing 12! in terms of 7! , allowing for easier calculations.

• Details on how to break down factorial calculations to simplify the expression further.

Simplifying Factorials

• Suggestion to use calculators but emphasizes learning manual simplification methods for evaluations without technology.

• Introduces an alternative method of simplification by canceling out common factors between numerator and denominator.

Final Calculation for Men

• After simplifications, it’s noted that there are 792 different ways to combine the selected men into groups of five.

• Reiterates that this number represents all possible combinations of choosing 5 from 12 men.

Calculating Combinations for Women

• Transitioning to calculate combinations for women using the same combination formula where n = 10 .

• Highlights the importance of ensuring non-negative values when calculating factorial differences (e.g., n - r ).

Final Calculation for Women

• Similar steps are taken as with men; simplifying factorial expressions leads to finding combinations efficiently.

• Concludes that there are 252 different ways to choose groups of five from ten women.

Total Combinations

Combinatorial Selection in Groups

Understanding Group Combinations

• The discussion begins with a scenario where a group is formed from 12 men and 10 women, emphasizing that the selection process does not differentiate between genders when choosing members.

• To simplify the selection of 10 individuals from this combined group, the total number of participants (22) is considered, making it easier to calculate combinations.

Key Concepts in Combinations

• The speaker highlights the importance of identifying whether a problem involves combinations or permutations. In this case, they focus on simple combinations without repetition.

• It’s noted that in combination problems involving people, repetition is generally not allowed; thus, specific formulas are applied to ensure accurate calculations.

Practical Exercise for Application

• An exercise is presented regarding selecting a committee consisting of three members from three different groups: Group A (7 members), Group B (10 members), and Group C (9 members).

• Viewers are encouraged to solve this exercise independently, with the answer being revealed later as 32 possible combinations.

Analyzing Selection Criteria

• The instructor explains that order does not matter when selecting committee members; hence, it qualifies as a combination rather than a permutation.

• Examples illustrate how different selections yield identical groups. For instance, selecting students A, B, and C is equivalent regardless of their order.

Calculation Steps for Combinations

• The calculation for each group's combinations is detailed:

• For Group A: binom73 = 7!/3!(7-3)!

• Similar calculations are performed for Group B and C using factorial simplifications to arrive at their respective combination counts.

Final Results and Conclusion

• After calculating all three groups' combinations separately and multiplying them together yields a total of 352800 ways to form the committee.