How many arrangements of 5αs, 5βs and 5γs are there with at least one β and at least one γ between each successive pair of αs? To solve this first I selected 1β from 5β's. Then I selected 2α's from 5α's. Then I selected 1γ from 5γ's and then I selected 2α's from the remaining 3. After this I divided the result by (4!) because 4 of the 
α's are common.
Therefore, the answer I got is ${(5C_{1}*5C_{2}*5C_{1}*3C_{2})\over4!}$.
Is this the approach for these type of questions?
EDIT:
I previously assumed that only γ was restricted to be placed between the pair of α's and also solved it using the integer approach which clearly was inaccurate. Also as pointed out by @RossMillikan since we are talking about arrangements order does matter.
 A: Treat the $5$ alphas as dividers. Assuming that the betas and gammas can only be placed between alphas, viz. $|...|...|...|...|$,
place a $\beta$ and a $\gamma$ in each of the $4$ "compartments".
The remaining $\beta$ and $\gamma$ can be placed in any compartment, thus $4\cdot4 = 16$ ways

Added
OP had opined that combinations were intende, but @Ross Millikan has observed that the word "arrangement" suggests permutations, and this could well be so, although the word alone is not a definitive indication. However, here is a working for permutations. There are two possible patterns:
$|\beta\beta\gamma\gamma|\beta\gamma|\beta\gamma|\beta\gamma|:$[Choose large cell and permute ]$\times$[Permute in other cells] $=[\binom41\cdot\frac{4!}{2!2!}]\times[2^3]$
$|\beta\beta\gamma|\beta\gamma\gamma|\beta\gamma|\beta\gamma|:$[Choose large cells and permute ]$\times$[Permute in other cells] $=[\binom42\cdot(\frac{3!}{2!})^2]\times [2^2]$
Multiply and add up.
A: Let us suppose that there are only $4\ \beta$s and $4\ \gamma$s. Then the question is easy, because we need to have an arrangement with exactly one $\beta$ and one $\gamma$ between every $\alpha$, and each time those can be ordered one of two ways, giving a total of $2^4=16$ options. 
Now we can introduce the fifth $\beta$. This can go at the start or end, or in any of the $4\ \alpha$-to-$\alpha$ internal zones. If the internal zone is chosen, the permutation for that zone rises from $2$ options to $3$, giving the result of $2\cdot 2^4+ 4\cdot 3\cdot 2^3 = 32+96 = 128$
Now add the fifth $\gamma$. Following the same process, this generates a slew of cases:


*

*fifth $\beta$ and $\gamma$ together, end zones: $2\cdot 2^5 = 64$

*fifth $\beta$ and $\gamma$ together, internal zones: $4\cdot 6\cdot 2^3 = 192$

*fifth $\beta$ and $\gamma$ separate, end zones: $2\cdot 2^4 = 32$ 

*fifth $\beta$ and $\gamma$ separate, internal zones: $6\cdot 3\cdot 2^2 = 72$

*fifth $\beta$ and $\gamma$ separate, mixed zones: $2\cdot 2 \cdot 4\cdot 3\cdot 2^3 = 384$


Total is $64+192+32+72+384 = 744$ options.
As noted your calculation as currently expressed is not an integer, so cannot be right. I don't really follow what you are attempting to do, so you'd need to add further explanation to get sensible feedback. 
