Here, n(total) = 400

n(chess ) = 18, n(TT) = 20, n (carom) = 27.

n(chess & TT)=7, n(TT & carom)= 12 and n(Chess & Carom)=?

n(chess & TT & carom)=4

We have, 

n(total) = n(chess) + n(TT) + n (carom)-n(chess & TT) - n(TT & carom)- n(chess & carom) +

n(chess & TT & carom)

$\Rightarrow$ 40=18+20+27-7-12-n(chess & carom) + 4

n(chess & carom) = 50 - 40 = 10

So, total number of students who play both chess & carom is = 10.

But there are 4 students who play all the three games. So, the number of students who play chess and carom but not table tennis = 10-4=6. ans.