A group of Construction students must choose their specialist options from the following list: Bricklaying, Damp-proofing, Drainage, Flooring, Joinery, Plastering, Roofing. Each student must choose three options. The available options may be combined with each other in any way, with the exception of the restrictions that Damp-proofing and Drainage together may not be combined with Bricklaying or Plastering because of timetable constraints, and that students choosing Joinery must also choose Flooring. How many possible combinations including Drainage could the students choose from?
A) 7
B) 8
C) 9
D) 10
E) 11

9

Step-by-step explanation:

Since drainage is to be included in the possible combinations, we can therefore determined the number from the remaining 6 options which can be in combination with drainage

(6 2) = 15 pairs.

Now there are restrictions placed, where we have no pairing of damp roofing with bricklaying or plastering. This removes two pairs, so we are left with 13pairs.

Another restrictions is that students choosing Joinery must also choose Flooring. This gives the three options: joinery, flooring and drained with no pairing with the remaining four pairs this removing another four pair.

So we are left with 9 pairs in total.

y-y1=4(x-x1)

step-by-step explanation:

this is point-slope form since you only have 4 as the slope it goes in place of m

hope this : )