Supposedly really hard problem involving combinations This problem gives 7 (max) out of 100 points for a college entrance exams. Seems odd because it looks easy to me, although my combinations are not too good.

There are $10$ people forming a commission. $2$ of them are students from different colleges. The commission is composed of $6$ members and if one of the students is in it the other must be as well. How many commissions like these can there be?

The answer is

 154

Any help is appreciated
 A: Assuming that either both or neither student must be in the commission, there are $$
\binom{8}{6}+\binom{8}{4} = 98$$
ways to form the commission.
Obviously this problem is not too easy, but it is a horrible problem for a big part of an entrance exam since most of the difficulty would be in guessing which of two ambiguous meanings the problem poser intended!
A: As @paw88789 pointed out in the comments, to obtain the answer $154$, we must interpret the question to mean if student $A$ is selected, then student $B$ must also be selected rather than student $A$ is selected if and only if student $B$ is selected.
Interpretation:  If student $A$ is selected, then student $B$ must also be selected. 
There are two possibilities.  Either student $A$ is selected or she is not.  
Case 1:  Student $A$ is selected. 
We know that if student $A$ is selected, then student $B$ must also be selected.  Hence, both students are selected, and the other four people must be selected from the other eight people available to serve on the commission.  Hence, the number of selections that include student $A$ is 
$$\binom{2}{2}\binom{8}{4}$$
Case 2:  Student $A$ is not selected.
The six members of the commission are selected from the other nine available people, so the number of selections that do not include student $A$ is 
$$\binom{9}{6}$$
Total:  Since the two cases are mutually disjoint, the number of ways of selecting a commission of six people given the restriction that if student $A$ is selected, then student $B$ must also be selected is 
$$\binom{2}{2}\binom{8}{4} + \binom{9}{6} = 1 \cdot 70 + 84 = 154$$ 
Interpretation: Student $A$ is selected if and only if student $B$ is selected.
This means that either both students are selected for the commission or neither is.
Case 1:  Both students are selected for the commission.
Since both students are selected, four of the other available people must be selected, so there are
$$\binom{2}{2}\binom{8}{4}$$
ways to select a commission of six people that includes both students.
Case 2:  Neither student is selected.
The six commission members must be selected from the eight other available people, so there are 
$$\binom{8}{6}$$
ways to select a commission of six people if neither student is selected.
Total:  Since the cases are mutually disjoint, the number of ways the commission of six people can be selected given the restriction that student $A$ is selected if and only if student $B$ is selected is 
$$\binom{2}{2}\binom{8}{4} + \binom{8}{6} = 1 \cdot 70 + 28 = 98$$ 
