Permutation or combination for repeated chosen items? I have this problem:
Choosing 5 vertices from the picture at random, what are the odds of at least 4 of them belonging to the pyramid?

My book says:

Because we are asked the probability of at least 4 vertices belonging
  to the pyramid, we must consider two possibilities: 4 of the chosen
  vertices belong to the pyramid or 5 of the vertices belong to the
  pyramid. This probability is given by:
$$\frac{^5C_4*^4C_1+^5C_5}{^9C_5}=\frac{1}{6}$$



*

*But I don't understand how this is. Wouldn't it be more logical to do
$$(\frac{5}{9}*\frac{4}{8}*\frac{3}{7}*\frac{2}{6}*\frac{4}{5})+(\frac{5}{9}*\frac{4}{8}*\frac{3}{7}*\frac{2}{6}*\frac{1}{5})
   = \frac{1}{126}$$
?

*Why is it that repetitions are not being counted? Shouldn't they be
part of the probability?

*Does the problem assume that the same vertex can be chosen more than
once?

*Why is my solution wrong?

*Can you explain to me my book's solution?


There is a chance my book is wrong, or the problem has an error, is missing information or is not explained properly. That sort of thing happens all the time . I copied the problem exactly as it is in the book.
 A: The book's solution is correct.  Your approach is viable, but you made an error in its execution.
The author of your text solved the problem by considering which five-element subsets of the vertices contain at least four vertices of the pyramid.  That is, the author is making an unordered selection of the vertices.  Since there are nine vertices in the diagram, there are 
$$\binom{9}{5}$$
ways to select a five-element subset of the vertices.  Five of the nine vertices are vertices of the pyramid.  Thus, a selection of five vertices which contains at least four vertices of the pyramid either contains four of the five vertices of the pyramid and one of the other four vertices in the figure or it contains all five vertices of the pyramid.  Thus, the number of favorable selections is 
$$\binom{5}{4}\binom{4}{1} + \binom{5}{5}\binom{4}{0}$$
Dividing the number of favorable selections by the number of possible selections yields the desired probability of 
$$\frac{\dbinom{5}{4}\dbinom{4}{1} + \dbinom{5}{5}\dbinom{4}{0}}{\dbinom{9}{5}} = \frac{1}{6}$$
You attempted to solve the problem by making ordered selections of five of the nine vertices.  You handled the case in which all five of the selected vertices are vertices of the pyramid correctly.  However, in handling the case in which exactly four of the five selected vertices are vertices of the pyramid, you failed to take the order of selection into account.  Your calculation for the probability of choosing four of the five vertices of the pyramid and one of the other four vertices in the figure
$$\frac{5}{9} \cdot \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{2}{6} \cdot \frac{4}{5}$$
is actually the probability of choosing four vertices of the pyramid, then choosing one of the other four vertices with your fifth selection.  However, the vertex that is not in the pyramid could have been chosen first, second, third, fourth, or fifth.  As you should check, if you add those five probabilities, you will find that the probability of choosing four vertices of the pyramid and one other vertex in the figure is five times what you calculated.  Observe that 
$$5 \cdot \frac{5}{9} \cdot \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{2}{6} \cdot \frac{4}{5} + \frac{5}{9} \cdot \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{2}{6} \cdot \frac{1}{5} = \frac{1}{6}$$
