One quibble here: the question isn't terribly clear about whether we want all the travelers to be in the same place, or whether we want all the travels to be in one predetermined place. It does seem likely to imply they're all in one place, whichever that place is, though, so I can live with it. Let me explain each situation, however.
For example, say y = 4, n = 3 and our destinations are Brazil, Madagascar, and Timbuktu. The odds of all four people arriving in Timbuktu are obviously lower than the odds of all four people simply traveling to one of any of the three countries.
If I want everyone to arrive in Timbuktu, the odds are (1/3) * (1/3) * (1/3) * (1/3), as we have four people, and each has a 1/3 chance of arriving in Timbuktu. The odds are thus (1/3)�, and generally speaking, (1/n)^y.
If I want everyone to arrive in any country, however, the odds are (1/3) * (1/3) * (1/3) * (1/3) * 3. The odds of the travelers arriving in any one country remain the same, but there are now three countries in which they could arrive, so I multiply the odds by 3. That gives me (1/3)³, or generally speaking, (1/n)^y * n, or (1/n)^(y-1).
One note about the dangers of posting exponents in text - as you've written the answers, Sana, it makes it look like both D and E are correct. I get that by E you mean (n/y)^n, which is not the correct answer, but E could also read as n/(y^n) ... which is the correct answer. For all positive n and y, n/(y^n) = (1/n)^(y-1). Here's an image demonstrating the algebra:
