Probability problem to kill some time (in Off-topic)

In an MVA test area there are only 4 chairs in a row available. At one instance, 5 men and 5 women appear for the test. If the test supervisor needs to place these people such that no two men or women are sitting next to each other, how many possible ways can these 10 people be arranged on 4 chairs?

800

1600

Jib, you're alive! Re-joining us? ;)

Unless I'm mistaken (which I could be at 3:30am), you were right the first time, deifeln.

There are two possible arrangements: MWMW or WMWM.
Take the first case: there are five men to choose from, then five women, then four men, then four women. Thus there are 5*5*4*4 = 400 combinations.

The other case is calculated the same way, with 400 combinations, for a total of 800 possible distributions of those 10 people.

By the way, the terms you want are "combinations", "permutations", "arrangements" or "distributions", and not "probability".

With probability, you'd be interested in the percent chance that a given man is sitting next to a given woman, for example.