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# Laura favorite clothes are five sweaters, four skirts, three jackets, two pair of shoes and one belt.

How many ways can Laura wear her favorite clothes without repeating the same exact outfit?

### 3 Answers by Expert Tutors

Steven L. | Professional Meteorologist Here to HelpProfessional Meteorologist Here to Help
5.0 5.0 (11 lesson ratings) (11)
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So this question asks how many outfits (consisting of one item of each of the five groups) can we make without repeating. To best explain this question, lets simplify it for a moment and say that Laura wants to know the combinations of only the 3 jackets, 2 shoes, and belt she can wear without repeating.

The three jackets will be represented as A1, A2, and A3,
The two shoes will be represented as B1 and B2,
and the Belt will be C1

Since we can only have one from each group, these are the only combinations possible:
A1 B1 C1
A1 B2 C1
A2 B1 C1
A2 B2 C1
A3 B1 C1
A3 B2 C1
So there are six possible combinations of just these three items. Mathematically, we could express this as 3 X 2 X 1 which is also six.

Another way of thinking about this is that we have three articles of clothing to select to start. Once we select a jacket, shoes, or belt we have two articles of clothing left to select. Once we select one of those, we have one article of clothing left to pick. So to find the number of combinations can we can multiply n by n-1 until we reach 0 or 3 X 2 X 1.

In the math world this can be represented as a factorial or 3! which is an easy way to represent 3 X 2 X 1.

We can do this for any number (n) articles of clothing. So let's go back to our original problem:

Laura needs to select outfits from 5 groups of clothing. It doesn't matter which order she picks the clothes in so the number of combinations would be 5! since n=5. If you listed every possibility out like I did above, you would indeed get the same answer as 5 X 4 X 3 X 2 X 1. If n=2 then you would do 2! or 2 X 1 which is 2. That would be asking now many combinations of the two shoes and belt can be made without repeating or (B1 C1) AND (B2 C1).

1! = 1 (Only 1 Possible Combo of Belts) = 1
2! = 2 (Only 2 Possible Combos of Shoes and Belts) = 2 X 1
3! = 6 (6 Combos of Jackets, Shoes, and Belts) = 3 X 2 X 1
4! = 24 (Number of Combos of Skirts, Jackets, Shoes, and Belts) = 4 X 3 X 2 X 1

So what is 5! or the number of combs of Sweaters, Skirts, Jackets, Shoes, and Belts?

Hope this helps!
John M. | John - Algebra TutorJohn - Algebra Tutor
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If Laura picks 1 of each type of clothing then she has the following
5 choices of sweater
4 choices of skirt
3 choices of jacket
2 choices of shoes
1 choice of belt

We don't care what order she wears the clothes in, but we don't want repetitions of an entire outfit.

Laura could wear the same skirt, jacket, shoes and belt with all 5 sweaters.
She could wear the same sweater, jacket, shoes and belt with all 4 skirts.

Following this method of thinking, we see that the number of possible outfits is
5 x 4 x 3 x 2 x 1 = 120.

Caitrin E. | Caclulus, Comp Sci, & Physics for visual & hands-on learnersCaclulus, Comp Sci, & Physics for visual...
4.9 4.9 (159 lesson ratings) (159)
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Hi Mitch,

It depends if you assume that Laura wears one item from each of these  categories (sweater, skirt, jackets, pairs of shoes, and belt) every day. If she does wear one item from each category she could mix and match like so:

(any 1 of the 5 sweaters) and (any one of 4 skirts) and (any one of 3 jackets) and (any one of 2 pairs of shoes) and (her 1 belt)

which comes to:

(5 sweater choices) * (4 skirt choices) * (3 jacket choices) * (2 shoe choices) * (1 belt choice) = 120 outfits

Here's the derivation:

Keep in mind that even if she keeps every other item of clothing the same but changes her shoes (or her jacket, or her sweater, or her skirt) it counts as a different outfit. So, if she has A combinations among her skirts, jackets, shoes, and belts, then she has 5*A outfits all together when she mixes and matches with different sweaters. If she has B combinations among jackets, shoes, and belts, then she has 5*4*B outfits by varying sweaters and skirts. If she has C combinations among shoes and belts, then she has 5*4*3*C outfits by varying sweaters, skirts, and jackets. Letting everything vary, we get: 5*4*3*2*1 = 120 outfits.

But be careful! *** If Laura does not have to wear a jacket or belt every day, then the answer is different! *** (It's probably safe to assume that she does have to wear a sweater, a skirt, and shoes every day, unless Laura's real name is Miley Cyrus or Nikki Minaj.) If Laura does not have to wear a jacket or belt every day, then she really has FOUR jacket choices (any of her three jackets + no jacket) and TWO belt choices (on or off). In this case she has:

(5 sweater choices) * (4 skirt choices) * (4 jacket choices) * (2 shoe choices) * (2 belt choice) = 320 outfits! Aah, fashion.

If this is a homework problem, you might want to submit both answers, just in case, with a clear explanation for each. (In your own words!)

Hope this helps,
Caitrin