The correct option is A 126
Let x1,x2,x3,x4,x5 be the fruits recieved by five people then
x1+x2+x3+x4+x5=20, x1,x2,x3,x4,x5≥3
Let X1=x1−3,X2=x2−3,X3=x3−3,X4=x4−3,X5=x5−3
So, X1+X2+X3+X4+X5=20−(3+3+3+3+3)=5, X1,X2,X3,X4,X5≥0
The number of non negative integral solutions is =5+5−1C5−1=9C4=126