Probability - Why is my answer wrong? Each item in a computer parts catalogue is given a unique code consisting of two distinct
uppercase letters followed by four distinct digits. For example, the code for a particular
keyboard is XY1702.
Suppose a sales clerk wants to order a particular item, and knows that the item code
begins with the letters DE, and that the digits 8 and 6 occur in the code. How many
different item codes meet this description?
Ans- Since DE is known and there is an occurrence of 8,6, then no. of ways two distinct numbers can be selected is 8P2. And no. of ways 4 distinct numbers can be arranged are 4!
Therefore, possible ways - 8P2 X 4! = 8!/(8-2)! X 4! = 1344
Why is my answer wrong?
Thanks all so much.
 A: We can select the $2$ missing digits from the $8$ candidates in $\binom{8}{2}$ ways. Then these two digits, along with $8$ and $6$, can be arranged in $4!$ ways, for a total of $\binom{8}{2}4!$.
Remark: You did something similar, but used $(8)(7)$ instead of $\binom{8}{2}$, so double-counted.
If you really want to use your technique, the places for the two known numbers can be chosen in $\binom{4}{2}$ ways. Then the numbers can be inserted in $2$ ways. And now the rest can be filled in $(8)(7)$ ways. 
A: First pick two locations for 8 and 6; there are four slots, so there are $4\cdot 3$ ways to do this.  Now you have something like  this
DE8_6_

You can populate the two slots with the remaining digits.  This can be done $8\cdot 7$ ways.  So there are 672 ways to do this.
A: Unique Code consisting of, in sequence, two distinct uppercase letters followed by four distinct digits ("choosing without replacement").
The code begins with DE. So we can disregard that. We're looking for the number of possible permutations of four digits including $8$ and $6$ each only once. There are ten digits to begin with, and having used up $8$ and $6$, we have $8$ possible digits. So there are ${8 \choose 2}$ possible combinations. Then we have four digits, which can be arranged in $4!$ ways: So we have ${8 \choose 2}\cdot 4!$ possible combinations.
