Five friends (including Omar and Sarah) go to a party and leave their jackets at the entrance. On their way out, they each pick one of the 5 jackets randomly (without replacement)
What is the probability both Omar and Sarah don’t end up with their own jacket?
I can't see how the answer given here makes sense. Clearly, I'm being a bit dense. The answer given is 13/20...
On
Using combinatorics, we can consider permutations of either jackets or people.
For example, we can ask which jacket Omar took and which jacket Sarah took. There are $5 \cdot 4 = 20$ different ways for Omar, and then Sarah, each to take one of the five jackets without replacement.
Now consider the following cases:
Both Omar and Sarah took jackets belonging to one of the other three friends. This can happen in $3 \cdot 2 = 6$ ways (ways to permute $2$ items selected from $3$ without replacement).
Omar takes Sarah's jacket, and Sarah takes one of the other three. This can happen in $3$ ways.
Sarah takes Omar's jacket, and Omar takes one of the other three. This can happen in $3$ ways.
Omar takes Sarah's jacket and Sarah takes Omar's. This can happen in only $1$ way.
Adding up the cases, $6 + 3 + 3 + 1 = 13.$ So the probability is $\frac{13}{20}.$
We could combine case 2 with case 4, or case 3 with case 4, but not both. I chose to leave them as three separate cases for symmetry.
Note that if we only had to deal with case 1, we could use combinations, where the order of the selected items does not matter; Sarah and Omar could swap jackets and still neither has their own. But in cases 2 and 3 the order definitely does matter; if they swap, at least one will get their own jacket back.
On
Use the complement event. The probability of Omar and Sarah not getting their jackets is equal to $1$ minus probability of at least one of them getting their jackets. Using inclusion-exclusion: $$1-\frac{4!+4!-3!}{5!}=\frac{120-42}{120}=\frac{78}{120}=\frac{13}{20}.$$ Interpretation:
Case 1: Omar gets his jacket. Then there are $4!$ ways for other four friends to get the rest jackets.
Case 2: Sarah gets her jacket. Then there are $4!$ ways for other four friends to get the rest jackets.
Case 3: Both Omar and Sarah get their jackets. Then there are $3!$ ways for other three friends to get the rest jackets.
Total number of ways for five friends to get the jackets is $5!$.
If Omar takes Sarah's jacket, with probability $\frac{1}{5}$, then the probability that they both get the wrong jackets is $\frac{1}{5}$.
If Omar takes one of the other three wrong jackets, with probability $\frac{3}{5}$, then the probability that they both get the wrong jacket is $\frac{3}{5}\cdot\frac{3}{4} = \frac{9}{20}$.
$$\frac{1}{5} + \frac{9}{20} = \frac{13}{20}.$$