Solving South America Combinations Problem: 210 Possibilities

[baywatch123]

Question: You are traveling to South America in two weeks. A friend bought you 5 watches and 6 pairs of sunglasses. You can bring at least 2 watches and at least 1 pair of sunglasses, and can only bring 4 items (so you don't lose all of them). How many combinations of watches and sunglasses can you have?

This is my thought process:
First, 5C2 (for the at least 2) = 10
Second, 6C2 (to fill remaining last 2 "spots") = 15
Multiply subsets = 150.

5C3 (since it was at least 2) = 10
6C1 (to fill in last spot) = 6
Multiply subset = 60

Add 150+60 = 210 possible combinations...

is this correct? is my thought process right. with the limit of "4 items" i feel like i didn't need to add "at least 1 pair of sunglasses"

what if the question instead stated, "you only want to bring 2 watches and 2 sunglasses..." then would it just be
5C2 = 10
6C2 = 15
and then multiply? to get 150 combinations?

thank you

 

[baywatch123]

sorry if i didn't follow the preformatted thread

 

[Orodruin]

It looks reasonable to me.

However, you did add the information on "at least 1 pair of sunglasses". If not you would also have had the option of taking 4 watches.

The following reasoning which gives another result would be wrong, can you tell why?
I will bring at least two watches and one pair of sunglasses so let me start by picking these. That would give me 5C2 = 10 and 6C1 = 6 options, respectively. There now remains 8 objects of which I can pick one for 8C1 = 8 options. Multiplying together gives 10*6*8 = 480 combinations.

 

[haruspex]

Duplicate of https://www.physicsforums.com/showthread.php?t=766202

 

[baywatch123]

Duplicate of https://www.physicsforums.com/showthread.php?t=766202

yeah sorry, i didn't mean to. didn't know where to ask until i started scoping around and found that this was the right place!

my apologies!

 

[baywatch123]

It looks reasonable to me.

However, you did add the information on "at least 1 pair of sunglasses". If not you would also have had the option of taking 4 watches.

The following reasoning which gives another result would be wrong, can you tell why?
I will bring at least two watches and one pair of sunglasses so let me start by picking these. That would give me 5C2 = 10 and 6C1 = 6 options, respectively. There now remains 8 objects of which I can pick one for 8C1 = 8 options. Multiplying together gives 10*6*8 = 480 combinations.

ah okay so the 8C1 is to fill in the last "remaining" spot

 

[Orodruin]

ah okay so the 8C1 is to fill in the last "remaining" spot

No, as I told you your first reasoning was correct. I added this as an example of how people get it wrong and asked if you can tell why it is wrong. (It is obviously double counting some combinations since the result was larger. But why does it overcount?)

 

[baywatch123]

No, as I told you your first reasoning was correct. I added this as an example of how people get it wrong and asked if you can tell why it is wrong. (It is obviously double counting some combinations since the result was larger. But why does it overcount?)

yes sorry i was referring to your example. and it would over count because now we might count in the "already chosen" watches and sunglasses

 

[Orodruin]

Yes, some of the combinations where a particular watch is chosen in the last step will be equivalent to combinations where it was chosen in the first and so on. Therefore, that method overcounts.

 

[haruspex]

yeah sorry, i didn't mean to. didn't know where to ask until i started scoping around and found that this was the right place!

my apologies!

As I pointed out on the other thread, it is not clear that you have to take four items. Do you?