Solution:
Given one-one function
f:{a,b,c,d}→{0,1,2,...10}
‌ and ‌2f(a)−f(b)+3f(c)+f(d)=0
⇒3f(c)+2f(a)+f(d)=f(b)
Case I:
(1) Now let f(c)=0 and f(a)=1 then
3×0+2×1+f(d)=f(b)
⇒2+f(d)=f(b)
Now possible value of f(d)=2,3,4,5,6,7, and 8 .
f(d) can't be 9 and 10 as if f(d)=9 or 10 then f(b)=2+9=11 or f(b)=2+10=12, which is not possible as here any function's maximum value can be 10 .
∴ Total possible functions when f(c)=0 and f(a)=1 are =7
(2) When f(c)=0 and f(a)=2 then
3×0+2×2+f(d)=f(b)
⇒4+f(d)=f(b)
∴ possible value of f(d)=1,3,4,5,6
∴ Total possible functions in this case =5
(3) When f(c)=0 and f(a)=3 then
3×0+2×3+f(d)=f(b)
⇒6+f(d)=f(b)
∴ Possible value of f(d)=1,2,4
∴ Total possible functions in this case =3
(4) When f(c)=0 and f(a)=4 then
3×0+2×4+f(d)=f(b)
⇒8+f(d)=f(b)
∴ Possible value of f(d)=1,2
∴ Total possible functions in this case =2
(5) When f(c)=0 and f(a)=5 then
3×0+2×5+f(d)=f(b)
⇒10+f(d)=f(b)
Possible value of f(d) can be 0 but f(c) is already zero. So, no value to f(d) can satisfy.
∴ No function is possible in this case.
∴ Total possible functions when f(c)=0 and f(a)=1,2,3 and 4 are =7+5+3+2=17
Case II:
(1) When f(c)=1 and f(a)=0 then
3×1+2×0+f(d)=f(b)
⇒3+f(d)=f(b)
∴ Possible value of f(d)=2,3,4,5,6,7
∴ Total possible functions in this case =6
(2) When f(c)=1 and f(a)=2 then
3×1+2×2+f(d)=f(b)
⇒7+f(d)=f(b)
∴ Possible value of f(d)=0,3
∴ Total possible functions in this case =2
(3) When f(c)=1 and f(a)=3 then
3×1+2×3+f(d)=f(b)
⇒9+f(d)=f(b)
∴ Possible value of f(d)=0
∴ Total possible functions in this case =1
∴ Total possible functions when f(c)=1 and f(a)=0,2 and 3 are =6+2+1=9
Case III:
(1) When f(c)=2 and f(a)=0 then
3×2+2×0+f(d)=f(b)
⇒6+f(d)=f(b)
∴ Possible values of f(d)=1,3,4
∴ Total possible functions in this case =3
(2) When f(c)=2 and f(a)=1 then,
3×2+2×1+f(d)=f(b)
⇒8+f(d)=f(b)
∴ Possible values of f(d)=0
∴ Total possible function in this case =1
∴ Total possible functions when f(c)=2 and f(a)=0,1 are =3+1=4
Case IV:
(1) When f(c)=3 and f(a)=0 then
3×3+2×0+f(d)=f(b)
⇒9+f(d)=f(b)
∴ Possible values of f(d)=1
∴ Total one-one functions from four cases
=17+9+4+1=31
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