A person has 5 new shirts and 3 new pants. If an outfit consists of a shirt and a pant, how many distinct outfits can the person have?

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A person has 5 new shirts and 3 new pants. If an outfit consists of a shirt and a pant, how many distinct outfits can the person have?

Mathematics
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15
why is the solution to this 5 * 3? why doesn't it use combination?
one pair goes with 5 different shirts to make 5 outfits while another pants pair makes another 5 outfits while the last pants pair makes 5 more outfits which im pretty sure leads to 5+5+5=15 if there is another way you are suppose to do ithe problem i don't know it

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It is order-dependent. Since each shirt-pants combination is unique, then we have that it is simply the amount of ways in which one can pick any numbering of shirts with any numbering of pants, and have a different combination in each case. I.e. Pants 1, Shirt 2 is a different combination than Pants 2, Shirt 1.
yes....but why is it 5 * 3 and not combination?
the question is the number of ways you can pick...so combination should be used right?
We use combinations on order-independent events. E.g. cards. Saying we have a hand of cards 1, 2, 3 is the same as saying we have a hand of 3, 1, 2, as all cards have the same value, it's simply a permutation of the other. Remember that a combination \({n \choose r}\)is an equivalence to: "The number of ways one can pick \(r\) elements from a set of length \(n\), such that no two sets of \(r\) elements are permutations of one-another."
yes...and this doesn't have an order....
it's just selection....
Just to butt in here, but it you can use combinations. You choose 1 shirt from 5, to get \(\binom{5}{1}\), and 1 pair of pants from 3, to get \(\binom{3}{1}\). Multiply it together, you get \[\binom{5}{1}\cdot\binom{3}{1}=5\cdot3=15\]
that's another thing i don't get...why multiply? why not add?
Well, yes, it's possible, just not necessary. It can be expressed as an identity of combinations, since we know that any one element will never be a permutation of another unless it's the same, single, element.
..but it does answer my question...
The way I was recently taught to think about it, was by using statements "and then" and "or you could." When you use the "and then" you multiply, and with the "or you could" you add. In this case, you choose a shirt, "and then" choose a pair of pants. However, this doesn't precisely answer your question either.
And, we multiply since, if we imagine a set of sets counting each possibility (in this case), the Cartesian product is their total outcome set. E.g. Let's say we have set \(S\) of shirts and \(P\) of pants, then, the total set of combinations is: \(S\times P\), whose cardinality is: \[ |S\times P|=|S||P| \]Where: \[ S\cap P=\emptyset \]
Much of this is very well explained an proven in any decent Probability Theory book/textbook. There's a few online from professors, who simply put them up.
actually...it does @KingGeorge ....does that always apply?
@LolWolf im not really comfortable with many math symbols....
Wait, in my answer or for the book/text?
Sorry, dealing with some other things now as well. But this can apply to every counting problem I've successfully completed so far. Just a couple weeks ago, one of my teachers recommended a certain method of counting. Give me a minute, and I'll explain it.
@LolWolf what you just said
First, to count something, you're basically counting sets. To count it, we can create a general set with the properties we want, and while we do it, count how many ways to do each step. I'll use this problem as an example. First up, we construct the set. Step 1: Pick a single shirt from a set of 5 shirts. Choices: \(\binom51\) "And then" Step 2: Pick a single pair of pants from 3 pairs. Choices: \(\binom31\). Hence, you have \(\binom{5}{1}\binom31\) ways to dress yourself today.
All right, so the "Cartesian product" is when we take an element of each of two sets and concatenate these two to form one more set, which itself becomes an element of our new set, and we do this for every item in each of the two initial sets. Jeez, that's quite a bit of "sets," but it looks more like this: Let's say we have \(S\) and an element \(a\) in it, and another set \(T\) with an element \(b\) in it, then the set \(\{a,b\}\) is an element of the Cartesian product, yet, neither \(a\) nor \(b\) are, individually. And, sorry for the unnecessary restriction: \(S\cap P=\emptyset\), I was thinking of something else. As for the other part: the cardinality of a set \(S\) (its length) is denoted \(|S|\).
sorry @LolWolf but i really am not a mathematician....that's too complicated for me....
If you want a slightly larger example, we can count the number of ways we can order a deck of 52 cards using this method.
would it turn out as 52!
\[\begin{array}{|c|c|c|} \hline \text{Step}&\text{Construction} & \text{Choices} \\\hline \\1&\text{Choose a first card}&52\\ \hline \\2&\text{Choose a second card}&51 \\\hline \\\vdots&\vdots&\vdots\\\hline \\52&\text{Choose a 52nd card}&1\\ \hline \\ \end{array}\] These are all "and then" statements, so you multiply everything together, and you get \(52\cdot51\cdot50\cdot...\cdot2\cdot1=52!\)
wow
oh yes,.... i see what you mean by and then now

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