Finally, that leaves us with 10 athletes for the bronze medal podium.The math looks like this:12 x 11 x 10 = 1320You might notice that this is the fundamental counting principle.
There is an exception to this rule that I’ll talk about in the next section.
Big Idea: If you are forming a group from a larger group and the placement within the smaller group is important, then you want to use permutations.
That is, either you are in the committee or out of the committee (there are no gold medalists here! Therefore, we have to set up one combination for boys and one for girls. This gives us: 3C2 = 3 and 4C2 = 6 The final step is once we’ve figured out the combinations above, we have to use the fundamental counting principle and multiply the total number of possibilities in a committee, not add them: 3 x 6 = 18, answer (C).
)The next thing to notice with this problem is that of the original 9 students, 2 leave, one boy and one girl. Many students get stuck on this step and wonder, why don’t I add them.
The difficulty is in knowing exactly which one it is—combinations or permutations?
One way to think of it is to think of permutations as the number of arrangements or orderings within a fixed group.How many different ways can these three medals be handed out? Even though the top three spots for the sprinters forms a subgroup, it is the ordering within that subgroup that matters greatly, and is the difference between a gold, silver, and a bronze medal. Now, we have one athlete fewer—since one is already on the gold medal podium.An easy way to solve this question mathematically is to imagine that the dashes below are the podium upon which each sprinter will stand (albeit the dashes are at the same level):____ ____ ____ gold silver bronze To find out the number of different arrangement, ask yourself how many athletes can stand on the gold medal podium? So that gives us a total of 11 for the silver medal spot.Since, I am ‘choosing’ from a larger group, in this case two separate groups, I want to use combinations.Remember: once we’ve chosen the 2 boys or 3 girls, the position within the committee doesn’t matter. Combinations and permutations are the bane of many students.In other words, students have no difficult identifying whether a question is a combinations/permutations problem.For instance, if I’m a basketball coach and I want to find out how many distinct teams I can form based on a group of people, I want to use combinations.To make sure you understand this important distinction, here are three different scenarios.The third space can be filled by any of the 2 remaining letters and the final space must be filled by the one remaining letter.The total number of possible arrangements is therefore 4 × 3 × 2 × 1 = 4!