In how many ways can integers $a_1
In how many ways can $4$ integers $a_1<a_2<a_3<a_4$ be chosen from the integers $1,2,3,...,26$ such that $5 \le a_i - a_{i-1} \le 7$ for all $i = 2,3,4$?
I'm not sure what I'm missing in my line of thinking:
Since $5 \le a_i - a_{i-1} \le 7$, the difference between two consecutive integers chosen must be $5,6$ or $7$ (since everything's an integer). If I let $a_1 = 1$, then by incrementing the next integers by $7$ I see that the maximum I get is $22$. Therefore, the difference $a_i - a_{i-1}$ could be anything among $5,6,7$ for all $i$. Thus I have $3^3$ possibilities. 
Similarly, letting $a_1$ be $2,3,4$ and $5$ all yields $3^3$ possibilities. 
When I let $a_1 = 6$, however, if I keep incrementing by the biggest difference, $7$, the max. I get is $27$, which is $1$ more than the allowed maximum. That means I'm not allowed to use all three $7$'s for the difference but a maximum of two $7$'s. That gives me $3^3-1$ possibilities. 
When I let $a_1 = 7$, the 'maximum' I get is $28$, $2$ more than $26$. Therefore, I can only use a maximum of one $7$. Since there is one case where I can use all three $7$'s and six cases in which I use two 7's, I have $3^3 - (1+6)$ possibilities. 
For $a_1 = 8$ the 'maximum' $a_4$ is 29. So to fit in with $26$, I must not use any $7$'s at all. That gives me $2^3$ possibilities.
For $a_1 = 9$ I can get $a_4 = 30$ 'max'. To offset the difference of $4$, I cannot use 7's at all and one of the 6's. That gives me $4$ choices.
For $a_1 = 10$ I have $31$ 'max'. Now I can use only all 5's, or a maximum of one $6$ only. That's $4$ possibilities as well. 
Finally, for $a_1=11$, I can only have the differences be $5$ all the time. So 1 possibility.
So the sum gave me $198$. But the answer is $216$. Where did I go wrong?
 A: Let the consecutive increments be $x, y, z$. There are $27$ choices of $x, y, z \in \{5,6,7\}$. 
Given a fixed choice of $x, y, z$, the number of ways to select the $a_i$'s is $f(x,y,z) = 26 - (a_4 - a_1) = 26 - (x + y + z)$. 
Since $f(x,y,z)$ is linear, it is clear that its average value is $26 - 3 \times 6 = 8$.
Thus the answer is $27 \times 8 = 216$.
A: There is a mistake for $a_1 = 7$. The number of possibilities of having two $7$ isn’t $6$ but : $\binom{3}{2} = 3$.
Moreover, for $a_1 =8$, you are saying that there can’t be $7$, yet what about :
$a_1 = 8, a_2 = 8+7 = 15, a_3 = 20, a_4 = 25$ ? (Same argument for $a_1 = 9$)
A: You've just skipped some possibilities.  For example, you say that when $a_1=8,$ you can't use any $7'$s at all, but that's not so.  You can use $5,6,7$ or $5,5,7.$  It's probably easier if you consider that you can make a maximum of $18$ with the $3$ increments.
A: It is true that $a_1 = 1,2,3,4,5$ each give you $3^3$ possibilities.
Next, looking at $a_1 = 6$, we see that the possibility of $7,7,7$ is excluded, but every other possibility  is included, so this gives $26$ possibilities.
Looking at $a_1 = 7$, only those totaling to $20$ or more are not allowed, which means that $757$, for example, is permissible, but you have excluded it. Hence we exclude $776,767,677$ and $777$ to get $23$ possibilities, not $27 -(1+6)$, rather $27-(1+3)$.
For $a_1 = 8$ only those totalling to $19$ or more must be excluded. That is, the above possibilities, plus $757$, $775$,$577$,$667,676,766$. This gives $17$ in total.
For $a_1 = 9$ the above, plus all permutations of $765$, and $666$, are excluded, giving $17-7 = 10$.
For $a_1 = 10$ the above plus all permutations of $755$ and $665$ are excluded, giving $4$ possibilities.
For $a_1 = 11$ only $555$ remains.
The total? $135 + 26 + 23+17+10+4+1 = 216$.
