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

Namely: a→0, b→ 1, 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.