Question: Evening Out

Comment on Evening Out

Why can't we take this approach:
stage 1: 4C1
stage 2: 5C1
stage 3: 3C1

and, then add these 3: 4+5+3= ?

greenlight-admin's picture

IF it were the case that the couple can do EXACTLY ONE activity then your approach would be correct. However, we're told that the couple does one OF EACH. So, for example, some possible evening plans include:
- Restaurant #1, Movie #4 and Teahouse #2
- Restaurant #2, Movie #1 and Teahouse #3
- Restaurant #3, Movie #2 and Teahouse #2
- Restaurant #2, Movie #2 and Teahouse #2
- Restaurant #3, Movie #1 and Teahouse #1
- Restaurant #4, Movie #5 and Teahouse #3
etc

Does that help?

Cheers,
Brent

Couldn't you break it into the 3 stages and use combinations for each stage and then multiply 4 x 5 x 3 to get 60 (according to the Fundamental Counting Principle)?
greenlight-admin's picture

Yes, we can also use combinations for each stage.
However, since nC1 always equals n, we don't necessarily need to use combinations.

For example, there are 4 restaurants (call them A, B, C, and D).
So, we can select 1 restaurant in 4 ways (i.e., choose A or B or C or D)

Or we can use combinations (since the order in which we select 1 restaurant does not matter)
We can select 1 restaurant from 4 restaurants in 4C1 ways (= 4 ways)

We get the same results either way.

Cheers,
Brent

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