Probability & Statistics (38 of 62) Permutations and Combinations - Example 3 By Michel van Biezen

Summary:

1. Introduction: The lecturer introduces permutations and combinations, aiming to explore both concepts through a specific example involving five colors: blue, green, yellow, orange, and red.

2. Combinations for Different Spaces: a. For five colors in five spaces, combinations are calculated using the equation $�(�,�)=\frac{�!}{(�-�)!\times �!}$. In this case, $�(5,5)=\frac{5!}{(5-5)!\times 5!}=\frac{5!}{0!\times 5!}=1$. b. For four spaces, $�(5,4)=\frac{5!}{(5-4)!\times 4!}=5$. c. For three spaces, $�(5,3)=\frac{5!}{(5-3)!\times 3!}=10$. d. For two spaces, $�(5,2)=\frac{5!}{(5-2)!\times 2!}=10$. e. For one space, $�(5,1)=\frac{5!}{(5-1)!\times 1!}=5$.

3. Permutations for Different Spaces: a. For five colors in five spaces, permutations are calculated using the equation $�(�,�)=�!$. In this case, $�(5,5)=5!$. b. For four spaces, $�(5,4)=5!$. c. For three spaces, $�(5,3)=5\times 4\times 3$. d. For two spaces, $�(5,2)=5\times 4$. e. For one space, $�(5,1)=5$.

4. Observations: a. The number of combinations increases as the number of spaces decreases. b. The number of permutations remains the same for four and five spaces. c. For one space, the number of combinations and permutations are equal.

5. Insight: The example highlights the differences between permutations and combinations based on the arrangement of colors in various spaces, providing insight into their applications.

This summary captures the lecturer's explanation of permutations and combinations using the example of arranging colors in different spaces, elucidating the relationship between the number of spaces and the resulting combinations or permutations.