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## Comment on

Evening Out## Why can't we take this

stage 1: 4C1

stage 2: 5C1

stage 3: 3C1

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

## IF it were the case that the

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

## Yes, we can also use

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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