Combination problem approach when at least one of one type must be included Question:In how many ways can 6 be chosen from 4 officers and 8 privates to include at least 1 officer?
The correct answer is to consider all the cases in which at least 1 officer is chosen and then add them together.


*

*1 officer + 5 private

*2 officer + 4 private

*3 officer + 3 private

*4 officer + 2 private


But what is wrong with the following approach (it gives wrong answer)
1 officer out of 4 can be chosen in $^4C_1$ ways and remaining 5 people out of 11 can be chosen in $^{11}C_5$ ways, so the total number of ways of selecting 6 people in which at least 1 officer is present = $^4C_1 * ^{11}C_5$ 
 A: 
Question:In how many ways can 6 be chosen from 4 officers and 8 privates to include at least 1 officer?

The easiest way is to count the total and subtract the complement.
(Count of those with At Least 1 Officer) = (Count of All Combinations) - (Count those with No Officers)
$$\,^{12}C_6 - \,^{8}C_6 = 896$$
This will give the same answer as summing all the ways.
$$\,^{4}C_1 \,^{8}C_5 + \,^{4}C_2 \,^{8}C_4 + \,^{4}C_3 \,^{8}C_3 + \,^{4}C_4 \,^{8}C_2 = 896$$

$\,^4C_1 \,^{11}C_5$ is the way to select 1 officer as leader of a team of 6 when the rest of the team can be filled by either officers or privates.   When the first officer selected does not have a distinct placement, you run into double counting issues. 
You're counting the ways to select officer 1 to 4 and, for each one, counting the ways to fill the rest of the team, including those with the remaining officers. But it doesn't matter if you select officer 1 first and officer 2 among the rest, or officer 2 first and officer 1 among the rest.  Those are the same combinations.
