Create two different situations: one in which you use combinations and one in which you use permutations
please help me with this question or give me some sites too look at, I don't understand.
Permutation: the order matters
e.g. In how many was can 8runners finish first, second , and third
number of ways = 8*7*6 = 336 or 8!/5!
combination:
A teacher wants to choose 3 students from 8 to form a clean up crew. The order in which the 3 students are chosen does not matter.
number of ways = 8!/(3!5!) = 56
google "permutations vs combinations"
http://www.mathsisfun.com/combinatorics/combinations-permutations.html
Thank you very much too the both of you!!! Now I understand!!!
Sure! I'll provide you with two different situations - one where combinations are used and one where permutations are used.
Situation 1: Using Combinations
Let's imagine you are arranging a dinner party and want to choose a dessert for your guests. You have 5 different options: chocolate cake, apple pie, ice cream, cheesecake, and brownies. However, you want to serve a combination of any 3 desserts. In this situation, you'll be using combinations.
To find the number of combinations, you'll use the formula for combinations: C(n, r) = n! / (r!(n-r)!), where n is the total number of options and r is the desired number of options selected.
Using the formula, the number of combinations for choosing 3 desserts out of the 5 options would be:
C(5, 3) = 5! / (3!(5-3)!) = 5! / (3!2!) = (5x4x3x2x1) / [(3x2x1)(2x1)] = 10.
So, in this scenario, there are 10 different combinations of 3 desserts that you can serve to your guests.
Situation 2: Using Permutations
Now, let's consider a situation where you want to arrange a group of 4 people in a line for a photograph. You have 6 people to choose from. In this case, you'll be using permutations.
To find the number of permutations, you'll use the formula for permutations: P(n, r) = n! / (n-r)!, where n is the total number of options and r is the desired number of options chosen or arranged.
Using the formula, the number of permutations for arranging 4 people out of the 6 options would be:
P(6, 4) = 6! / (6-4)! = 6! / 2! = (6x5x4x3x2x1) / (2x1) = 360.
Here, there are 360 different permutations of arranging 4 people from the group of 6 for the photograph.
I hope this clarifies the difference between combinations and permutations and how they are used in different situations. If you need further assistance or want to explore more examples, feel free to ask!