I am a bit confused. When we use the multiplicative principle are we finding the number of permutations or combinations.
An example of using this principle is where I have $5$ shirts $3$ pairs of pants and $2$ pairs of shoes and I want to find out how many ways I can wear the ($1$ shirt, $1$ pair of pants and one pair of shoes). So we just multiply $5 \times 3 \times 2 = 30$ ways. So, is this the number of permutations or combinations?
Thank you 🙂
Best Answer
Your example is just saying: how many different unique outfits can I make if I wear one shirt, one pair of pants and one pair of shoes.
So you say that there are 3 things you need to do:
choose 1 shirt, choose 1 pair of pants and choose 1 pair of shoes.
Multiplicative principle: If I have $5$ different shirts, $3$ different pairs of pants and $2$ different pairs of shoes, then there are $5 \times 3 \times 2$ ways of taking $1$ shirt, $1$ pair of pants and $1$ pair of shoes. You choose $1$ of each and multiply them together because this suggests they happen at the same time.
Sum Principle: If I have $5$ different red shirts, $3$ different blue shirts and $2$ different black shirts, then I have $5+3+2 = 10$ different shirts.
The number of possible ordered samples of size $r$ taken from a set of $n$ objects is $n^r$ when sampling with replacement. (you can still choose from all $n$ objects each time for $r$ samples)
e.g. If I have a $3$ digit code and each digit can be any number $0,1,2,...,9$ (10 options). How many codes can I have?
Answer: $10 \times 10 \times 10 = 10^3 = 1000$
The number of permutations of $n$ distinct objects with replacement is $n!$
e.g. Say I have 5 different books and I need to place them all on a bookshelf in 5 places. How many ways can I do this?
So you can choose from $5$ books in the first place. You have $4$ books left. Now you can place $4$ books in the second place. Do you understand that we are not replacing the books. Once we place one our sample size decreases.
Answer: $5 \times 4 \times 3 \times 2 \times 1 = 5! = 120$
The number of permutations of $n$ distinct objects when you only want to permute $r$ objects and not all $n$ but you are still taking them from $n$ objects without replacement:
$$^nP_r = \frac{n!}{(n-r)!}$$
E.g. Say I have 5 different colored ribbons. How many ways can I give 3 people ribbons such that each person gets only 1.
Answer: $5 \times 4 \times 3 = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 5 \times 4 \times 3$
Do you see what I'm getting at. Logically you can just say $5 \times 4 \times 3$ but when numbers get very large using the general formula is helpful. It allows you to cancel out all the unwanted numbers.
Now combinations are where you are getting the number of ways in which $r$ objects can be combined but the order is not important anymore:
$$^nC_r = {n \choose r} = \frac{n!}{r!(n-r)!}$$
Note that the difference between this and the permutation is that you divide the permutation further by $r!$
E.g. Say that we have $5$ different coloured ribbons as above but instead of placing them in an order we want to see how many ways we can choose $3$ ribbons.
The fact that order is not important means that if you choose B,R,G (blue, green, red) then it is the same as choosing B,G,R or G,B,R or G,R,B or R,G,B or R,B,G. Note that if we are choosing 3 then we need to divide by $3!=6$ because that is how we eliminate the need for order
Answer: ${5 \choose 3} = \frac{5!}{(5-3)!3!} = \frac{5 \times 4 \times 3}{3!} = 10$