Arrangements of $5$ $\alpha$s, $5$ $\beta$s and $5$ $\gamma$s with at least one $\beta$ and at least one $\gamma$ between each two $\alpha$s How many arrangements of $5$ $\alpha$s, five $\beta$s and five $\gamma$s are there with at least one $\beta$ and at least one $\gamma$ between each successive pair of $\alpha$s?
My attempt:
case 1: exactly $1$ $\beta$ and $1$ $\gamma$ b/w each pair of $\alpha$s $= 96$ ways
case 2: exactly $1$ $\beta$ and $2$ $\gamma$s b/w each pair of $\alpha$s $= 192$ ways
case 3: exactly $2$ $\beta$ and $1$ $\gamma$ b/w each pair of $\alpha$s $= 192$ ways
Don't know how to compute case 4. I mean I know but the answer is not matching.
case 4 would be exactly $2$ $\beta$s and $2$ $\gamma$s b/w each pair of $\alpha$s.
If anyone knows another best possible solution please provide or else just help me with case 4.
Thanks!
 A: You should describe your cases in a little different way (results are right). Also there are more cases.
Case 1. Between each consecutive pair of $\alpha$ there are one $\beta$ and one $\gamma$
Each pair $\beta,\gamma$ can be placed in two ways: $\beta\gamma$ or $\gamma\beta$
The last $\beta$ and $\gamma$ should be placed 'outside' all alphas - it can be done in $2\cdot 3=6$ ways
Thus there are 
$a_1 = 2^4\cdot 6 = 96$ 
ways 
Case 2. As above, but there is also exactly one pair of consecutive alphas with 2 betas ond one gamma
$\beta,\beta,\gamma$ can be arranged in $\binom{3}{1}=3$ ways: $\beta\beta\gamma$, $\beta\gamma\beta$, $\gamma\beta\beta$
Position of $\beta,\beta,\gamma$ can be selected in $\binom{4}{1}=4$ ways.
The last $\gamma$ can be placed in 2 ways.
Thus we have 
$a_2 = 2^3\cdot 3\cdot 4\cdot 2= 192$
Case 3. As above, but there is exactly one pair of alphas with 2 gammas ond one beta (instead of of 2 betas ond one gamma)
Calculations are analogous to these from case 2, so
$a_3 = a_2= 192$
Case 4: There is exactly one pair of consecutive alphas with two betas and two gammas between them
$$\beta,\beta,\gamma,\gamma$$ can be arranged in $\binom{4}{2}=6$ ways and can be placed in $\binom{4}{1}=4$ ways
$a_4=2^3\cdot 4\cdot 6 = 192$
Case 5: (not considered in your solution) There is exactly one pair of consecutive alphas with 2 gammas ond one beta and exactly one pair of consecutive alphas with 2 betas ond one gamma
$\beta,\beta,\gamma$ and $\gamma, \gamma, \beta$ can be placed in $4\cdot 3=12$ ways
$a_5 = 2^2 \cdot 3^2 \cdot 12 = 432$
A: Using Roman letters and making the $A's$ dividers for simplicity, note that
between two $A's$, you can have $BC$ ($2$ perms), $BBC/BCC$ ($3$ perms) or $BBCC$ ($6$ perms)
The basic patterns are:
$00|00|00|00|00|\to 2^5\times 2$ (for outermost $00$ at other end) $=64$
$0|00|00|00|00|0\to 2^5 = 32$
$0|000|00|00|00|\to 3\cdot2^4\times 4\times 2$ (placing $ 000\;$ and for outermost $0$ at other end) $=384$
$|0000|00|00|00|\to 6\cdot2^3\times 4$ (placing $0000$) $=192$
$|000|000|00|00|\to 3^2\cdot2^2\times (4\cdot3)$ (placing the two $000's$) $= 432$
Total $=64+32+384+192+432= \boxed{1104}$ 
