TITLE: Number of ways in which $4$ people can be selected out of $10$ people sitting in a row such that exactly two are consecutive
QUESTION [5 upvotes]: Number of ways in which $4$ people can be selected out of $10$ people sitting in a row such that exactly two are consecutive is?

My attempt:
I tried two approaches:

PIE: required ways = total ways - ways in which three are consecutive + 
ways in which four are consecutive. But I do not know how to calculate "ways in which three are consecutive", so I am stuck.
Direct: the problem is that:

If I select two people at the ends: - - 3 4 5 6 7 8 9 10 or 1 2 3 4 5 6 7 8 - -, then I have six more ways to select the third person, but am unsure about the fourth way.
If I select two people in the middle, like so 1 2 3 - - 6 7 8 9 10, then selecting the next two people is troublesome, as the seats 1 and 2 cannot be occupied.


Both these approaches seem cumbersome. I hope there's a simpler way. What is that?

REPLY [3 votes]: Direct method: let the people be $1,2,...,10$.
Consider two consecutives: 
$$\begin{align} &1,2: \ {7\choose 2}-{6\choose 1}=15 \\
&2,3: \ {6\choose 2}-{5\choose 1}=10\\
&3,4: \ {6\choose 2}-{4\choose 1}=11\\
&\vdots\\
&7,8: \ {6\choose 2}-{4\choose 1}=11\\
&8,9: \ {6\choose 2}-{5\choose 1}=10\\
&9,10: \ {7\choose 2}-{6\choose 1}=15.\end{align}$$
Hence, the sum is $105$.