Since the 1 particular tutor must be on the committee, we 
place that particular tutor on the committee and then the 
problem becomes to select only 1 tutor from the 5 remaining 
tutors and 5 pupils without those two BOTH being on the
committee. 

First we find the number of committees without regard to
whether or not those two pupils are BOTH on the committee.
Then we will subtract the number of committees which they 
are both on.

We can select the remaining tutor "5 tutors choose 1"  5C1 ways.
We can then select the the 5 pupils "10 pupils choose 5, 
or 10C5 ways.

That's (5C1)(10C5)

From that we must subtract the number of committees where those 
particular pupils are together on the committee:

We can select the remaining tutor "5 tutors choose 1"  5C1 ways.
We can then select the other 3 pupils besides those 2 in 
"8 pupils choose 3: or 8C3 ways. 

That's (5C1)(8C3) that we must subtract:

Answer: (5C1)(10C5)-(5C1)(8C3)=(5)(252)-(5)(56)=1260-280=980 ways.

Edwin