I am studying for Exam P, here is the question : Five married couples attend a l

Question

I am studying for Exam P, here is the question:

Five married couples attend a local dance. If partners are randomly assigned, what is the probability that fewer than three couples end up being paired?

*this is a combonation question, not a permutation as far as I know

Here is the Answer (in detail):

1. P(A) = 1 - P(A compliment), n(A compliment) = n(3) + n(4) + n(5)

2. n(3) = 5C3 = 10, n(4) = 0, n(5) = 1,

thus, P(A) = 1 - (10 + 0 + 1)/ 5!

= 1- (11/120) = 109/120

= 0.908 or approx. 91%

What I Know:

1. 5C3 = 10, when I type it in in my calculator, and it fits within the rule of combinations, thus n(3) = 10

2. 5C5 = 1, this is also true when I type it in my calculator, thus n(5) = 1

My question:

Why does n(4) = 0?

Explanation / Answer

n(4) refers to the possibility of exactly 4 couples being paired and one couple being unpaired. This is not possible, because if 4 couples are paired, then the balance one couple will also be paired by default, because there is no one else to pair with them. So if 4 couples are paired, then the 5th couple is also paired and we have the case of 5 paired couples, i. e. n(5).

So the counts for exactly 4 paired couples is 0, i. e., n(4) = 0

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