Combinatorics: What to Do with "Forbidden Choices"
David and Rachel are getting married. The extended family wants to have their picture taken, but David's father is unwilling to stand next to Rachel's mother. How many options does the photographer have to arrange the 10 family members in a row for the picture?
Incorrect.
[[snippet]]
Incorrect.
[[snippet]]
Correct.
[[snippet]]
There are $$10!$$ options to arrange 10 people in a row. Now, calculate the forbidden options. To do that, consider the two enemies as a single "person." So now there are $$9!$$ ways to arrange 9 people in a row.
We're not done. Since the two enemies have $$2!$$ ($$=2$$) internal arrangements between them (the father on the left, then the mother, and vice versa), the number of forbidden arrangements is $$9!\times 2!$$.
Subtract the forbidden choice from the total number of choices to get only the good options:
>$$\text{Good combinations} = 10! - (9!\times 2)$$
You can simplify this by factoring. Remember,
>$$10! = 10\times (9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times1)$$,
>$$\ \ \ \ = 10\times 9!$$
Use this to factor out $$9!$$:
>$$ 10! - (9!\times 2) = 10\times 9! - 2\times 9!$$
>>>$$ \ \ \ \ = 9! \times (10-2)$$
>>>$$ \ \ \ \ = 9!\times 8$$
$$9!$$
$$9\times 8!$$
$$8\times 9!$$
$$\frac{10!}{2!}$$
$$10!$$
