The correct option is C 14C5 if each person gets even number of things(excluding 0).
Let person Pi gets xi number of things such that :
x1+x2+x3+x4+x5+x6=30
Case I - If xi is odd or xi=2yi+1, where yi≥0, then
2(y1+y2+y3+y4+y5+y6)+6=30⇒y1+y2+y3+y4+y5+y6=12
Then the number of solutions
= 12+6−1C6−1= 17C5
Case II - If xi is even or xi=2yi,
where yi≥1(person getting 0 items is excluded), then
2(y1+y2+y3+y4+y5+y6)=30⇒y1+y2+y3+y4+y5+y6=15
Then the number of solutions
= 15−1C6−1= 14C5