Online Course Discussion Forum

MCIIA Combinatorics 1.21

 
 
Picture of Maximus Sheng
MCIIA Combinatorics 1.21
by Maximus Sheng - Saturday, 6 May 2023, 10:11 AM
 
I don’t get how you’re able to just multiply everything out.


I think the process should be like this:

4*4*3(case in which you pick pants) + 4*4*5(case in which you pick shorts) = 128

Case in which you pick only a hat: 128*2 = 256

Case in which you pick only a jacket: 128*1 = 128

Case in which you pick both: 128*2 = 256

Case in which you pick none: 128

256*3 - 768.

 
Picture of John Lensmire
Re: MCIIA Combinatorics 1.21
by John Lensmire - Monday, 8 May 2023, 12:52 PM
 

Problem (1.21 from MC II-A Combinatorics) for reference:

John has 2 hats, 4 shirts, 1 jacket, 3 pairs of pants, 5 pairs of shorts, and 4 pairs of shoes. Suppose an outfit always contains a shirt, legwear (either shorts or pants), and shoes. Further, an outfit may contain a hat and/or a jacket. How many outfits in total does John have?

First off, note that you do get the same answer each way, so it is okay to use cases, but they are not required here.

The key here is that actually NONE of the choices affect other choices. (This is different from one of the examples involving shorts vs pants). Thus, we are able to make all these choices independently from the others (and can use the Product Rule) to multiply out the number of outcomes).

I like thinking of a problem like this as "ordering an outfit from a menu". Note we can make our menu:

  • Shirt: Shirt A, ..., or Shirt D ($4$ choices)
  • Legwear: Shorts A, ..., Shorts E, Pants A, ..., Pants C ($8$ choices)
  • Shoes: Shoes A, ..., Shoes E ($5$ choices)
  • Hat: Hat A, Hat B, or No Hat ($3$ choices)
  • Jacket: Jacket A or No Jacket ($2$ choices)

Then, multiplying we get our answer $4\cdot 8\cdot 5\cdot 3\cdot 2 = 768$. Hope this helps!