f(a)=0,f(b)= 1,f(c)= 1
Namely: a→0, b→ 1, c→ 1.
The second type: 1+0= 1.
f(a)= 1,f(b)=0,f(c)= 1
The third type: 0+0=0
f(a)=0,f(b)=0,f(c)=0
Namely: a→0, b→0, c→0.
So the number f of mappings is 3.