# There are 10 men and 10 women. Alice (woman) and Brad (man) never stand next to each other.

There are 10 men and 10 women. Alice (woman) and Brad (man) never stand next to each other. Assume that the men and women alternate in a line, how many possible ways are there for the arrangement?

Without the condition of Alice and Bob not standing next to eachother, it would be 10! * 10! * 2. Now I need to subtract all the ways in which they could be stuck together, or 9! * 9! * 10 * 2. (I can elaborate on how I got this part if it is wrong)

So would the final answer just be 10!10!2 - 9!9!10*2 ??

• I think you meant $(10+10)!$ instead of $10!*10!$. – 79037662 Oct 4 '19 at 19:41
• @79037662 initially men can be in 10 spots, so 10!, and same for women so wouldn't it be multiplied – kmorg Oct 4 '19 at 19:43
• It certainly would be multiplied. @79037662 you missed the condition that the men and women alternate places. – JMoravitz Oct 4 '19 at 19:43
• @JMoravitz My mistake. – 79037662 Oct 4 '19 at 19:46
• Your answer is very close kmorg, and has good logic behind it. However, you seem to have only accounted for the situations where alice and brad occupy one of the groups $(\star~\star)(\star~\star)(\star~\star)\cdots$, that is, where the smaller index is odd but I think you missed situations like $\star(\star~\star)(\star~\star)\cdots$ where they were placed in adjacent pairs but not with one another, that is, where the smaller index position is an even number (where indexing starts at one) – JMoravitz Oct 4 '19 at 19:47

Pick the position of the leftmost member of alice and brad. There are $$19$$ options.
Then, pick whether it goes male-female-male-female-... or vice versa. $$2$$ options.
Finally from left to right for each spot designated for a male pick one of the remaining males to fill it and same for the females. $$9!\cdot 9!$$ options.
This gives a total of $$19\cdot 2\cdot 9!\cdot 9!$$ options for Alice and Brad to stand next to one another, not $$10\cdot 2\cdot 9!\cdot 9!$$ as you had calculated. You had effectively missed the cases where the leftmost of them was in an even index (with indices counting from one).
Continuing, this gives $$10!\cdot 10!-19\cdot 2\cdot 9!\cdot 9!$$ ways for them to not stand next to one another.