Determine Which technique to use for probability Why is my choice of solving the problem wrong?
A snack stand sells 5 different flavors of italian ice. How many combinations of 4 different flavors of Italian ice cream can Sharon buy from this stand?
My Attempt: $5\times4\times3\times\cdots$
Correct Anwser $5$ aka ${}_5C_4$. 
The yearbook staff must assign 4 distinct photographs, one photograph per page, to 4 different pages. How many different assignments of photographs are possible? 
My Attempt ${}_4C_1$. 
Correct anwser $24$.
 A: $n \choose r$ or $^nC_r$ is the number of ways of choosing $r$ objects from $n$ in any order - i.e. where it doesn't matter what order we choose them in.
In the ice cream example, we want to know how many combinations there are. But we don't care in what order the flavours occur. For example, say the flavours are chocolate, vanilla, strawberry, banana and caramel, then if I wasn't a fan of strawberry, I would choose


*

*Chocolate, Vanilla, Banana, Caramel

*Banana, Chocolate, Vanilla, Caramel


Both these combinations have the same flavours - i.e. no strawberry - and hence count as the same combination. In this case, we want to know the number of ways of choosing $4$ flavours from a list of $5$ in any order, so the answer is $^5C_4$.
In the second example, each photograph must be assigned to a page - and hence this time the order matters. For example, if the pictures are lettered A, B, C and D, then


*

*Page 1: A, Page 2: B, Page 3: C, Page 4: D

*Page 1: B, Page 2: A, Page 3: C, Page 4: D


are different orderings. In this case, we have:


*

*$4$ choices for page $1$

*$3$ choices for page $2$ - since after putting one photograph on the first page, we have $3$ left

*$2$ choices for page $3$

*$1$ choice for page $4$


Hence the total number of combinations is $4\times3\times2\times1 = 24$.
A: Combinations of 4 flavors are just sets of 4 flavors. Each such set is uniquely determined by the unique remaining flavor. Hence 5 of them.
Assignments of photographs are different if each page each photograph lands to is not the same. Hence 4 choices for the first page, then 3 for the second page, then 2 for the third, and the assignment is complete. Hence 4x3x2=24 of them.
There are $\displaystyle{n\choose k}$ combinations of $k$ flavors amongst $n$.
There are $\displaystyle\frac{n!}{(n-k)!}$ assignments of $k$ photographs amongst $n\geqslant k$, to $k$ pages.
A: Ok for the first one order doesn't matter obviously. There are 5 choices of those 5 you must choose 4. Thus 5C4.
For the second order matters since there are different positions  thus 4!=24.
