select sequence without "ABCD" There are an infinite number of coupons bearing one of the letters A,B,C,D. Find the number of ways for selecting $10$ of these coupons so that there shouldn't be sequence of "ABCD", which means after selecting a 'A' coupon and an 'B' and a "C" coupon you cannot select a 'D' coupon immediately.
I tried like first all possible combination equals $4^{10}$ and then considered sequence "ABCD" together by means $4$ consecutive places are fixed so,total available positions reduce to $7$. In $7$ ways we can place "ABCD" and then remaining $6$ places as selecting any number out of $4$, which equals $4^6$. But it does not work. Looking for kind help.
 A: As I said in the comments, you need to adjust for the double counting by adding back the sequences containg two copies of $ABCD$. You can place these two copies in $6$ different ways and choose the two remaining numbers, thus you have $6\cdot 4^2$ such sequences. Hence the total number should be
$$4^{10}-7\cdot 4^6+6\cdot 4^2.$$
A: I want to give you more powerful tool to solve these types of question . I learned it form @Markus Scheuer . (You can check over his answers to improve your combinatorics skills ).
Lets go over our question . Firstly , this question is more like recurrence relation question . I write these answer to give another perspective ,by the way  @b00n heT 's answer is more elegant and concise than my answer for this question.
Our method is called Goulden-Jackson method , it is a little detailed article , so i wont get in detail to explain it. Look at : Goulden-Jackson Cluster Method.
Our method says our alphabet is $V=\{A,B,C,D\}$ . Moreover , our bad word is $ABCD$. Now, $$A(z)=\frac{1}{1-dz-\text{weight}(\mathcal{C})}$$
$A(z)$ will be our generating function to find our recurrence relation. $d$ in $A(z)$ is equal to the number of elements in $V$.
$weight(C)= weight[(ABCD)]=-z^4$ , because there is not any overlapping in $ABCD$. You can find how to calculate it in given link.
Hence , $$A(z)=\frac{1}{1-4z + z^4}$$
It is the expansion of $A(z)$ , https://www.wolframalpha.com/input/?i=expanded+form+of+%281+%2F+%281-4x+%2Bx%5E4%29%29
You see  that the coefficient of $x^{10}$ is equal to $1020000$ , which is equal to @b00n heT 's answer
