9.6 Learning How to Count in Pre-Calculus
9.6 Learning How to Count in Pre-Calculus
First off know your number line.
For a reference:
To start off with some examples,
Ex. 1. If you have 4 different types of cheese to put on a pizza, and 4 different types of toppings, how many combinations can you create.
1 and 3 2 and 3 3 and 3 4 and 3
1 and 4 2 and 4 3 and 4 4 and 4
Ex. 1. If you have 4 different types of cheese to put on a pizza, and 4 different types of toppings, how many combinations can you create.
1 and 1 2 and 1 3 and 1 4 and 1
1 and 2 2 and 2 3 and 2 4 and 21 and 3 2 and 3 3 and 3 4 and 3
1 and 4 2 and 4 3 and 4 4 and 4
That creates 16 combinations. It could simply be solved by 4 · 4 = 16.
Ex. 2. If can wear 10 different wigs or 15 different natural hairstyles, how many hair styles can you choose from?
Well you could wear:
Wig #1 Wig #6 Hairstyle #1 Hairstyle #6 Hairstyle #11
Wig #2 Wig #7 Hairstyle #2 Hairstyle #7 Hairstyle #12
Wig #3 Wig #8 Hairstyle #3 Hairstyle #8 Hairstyle #13
Wig #4 Wig #9 Hairstyle #4 Hairstyle #9 Hairstyle #14
Wig #5 Wig #10 Hairstyle #5 Hairstyle #10 Hairstyle #15
And that is 25 different hairstyles which could be solved by 10 + 15 = 25.
If you would like, multiplication can be determined by "and" and addition can be determined with "or".
Ex. 2. If can wear 10 different wigs or 15 different natural hairstyles, how many hair styles can you choose from?
Well you could wear:
Wig #1 Wig #6 Hairstyle #1 Hairstyle #6 Hairstyle #11
Wig #2 Wig #7 Hairstyle #2 Hairstyle #7 Hairstyle #12
Wig #3 Wig #8 Hairstyle #3 Hairstyle #8 Hairstyle #13
Wig #4 Wig #9 Hairstyle #4 Hairstyle #9 Hairstyle #14
Wig #5 Wig #10 Hairstyle #5 Hairstyle #10 Hairstyle #15
And that is 25 different hairstyles which could be solved by 10 + 15 = 25.
If you would like, multiplication can be determined by "and" and addition can be determined with "or".
Permutations
A permutation contains different numbers which are assigned a position such as first, second or third.
This is when factorials come into place.
Ex 1. In the WCSR (World Championship of Snail Racers), there are 7 snails racing. What are the possible outcomes of the race.
So any of the 7 could win first, so 7.
Any of the remaining could win second, so 6.
Then the 5 remaining for the rest of the positions, so 5.
And so on for 4, 3, 2, 1.
You could then determine it is 7 · 6 · 5 · 4 · 3 · 2 · 1.
Which miraculously is 7!! (exclamation point at the end of a factorial, bam)
=5040
So there are 5040 possible outcomes of the race. Not including ties of course.
Ex 2. Now if the problem is getting tricky, such as it asks you for the possible outcomes for first, second and third.
But when they are trying to trick you, it simply means mind tricks, and you can out trick the mind tricks by tricking your mind into tricking the mind tricks into non-mind tricks, which is what the problem is asking you to do.
By following that, you can determine the same principle above, you can infer,
Any of the 7 could win first, so 7
Any of the 6 remaining could win second, so 6
And any of the 5 remaining could win third, so 5
That means simply 7 · 6 · 5
Which equals 210 possible combinations.
Where n is the number of snails, and r is the number of positions you are looking for.
Combinations
Combinations are similar to permutations in that they calculate the possible number of times something could occur.
The difference is that in combinations order does not matter.
In my opinion you could take this two different ways
1. Order does not matter, therefore ABC is different than CBA, because they are in different order and it doesn't matter, they both count.
2. Order does not matter, therefore even if ABC and CBA are in different orders, it doesn't matter, they don't count.
But apparently it is number 2.
So even if they are in different orders, the same combination of letters would not count.
Ex. 1. You have to type a paper only using the letters R S T L N E (anyone?), how many combinations of words could you make using each letter once for a 6 letter word (order does not matter)?
So we have 6 letters, how many word combinations can we make with 4 letters.
We get 15 possible word combinations.
Wrong, its 4.
Theres only 4 words, the rest aren't words. :/
Ex. 1. You have to type a paper only using the letters R S T L N E (anyone?), how many combinations of words could you make using each letter once for a 6 letter word (order does not matter)?
So we have 6 letters, how many word combinations can we make with 4 letters.
We get 15 possible word combinations.
Wrong, its 4.
Theres only 4 words, the rest aren't words. :/














































