Given one-one function

$$f:\{ a,b,c,d\} \to \{ 0,1,2,\,\,....\,\,10\} $$

and $$2f(a) - f(b) + 3f(c) + f(d) = 0$$

$$ \Rightarrow 3f(c) + 2f(a) + f(d) = f(b)$$

Case I:

(1) Now let $$f(c) = 0$$ and $$f(a) = 1$$ then

$$3 \times 0 + 2 \times 1 + f(d) = f(b)$$

$$ \Rightarrow 2 + f(d) = f(b)$$

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.

$$\therefore$$ Total possible functions when $$f(c) = 0$$ and $$f(a) = 1$$ are = 7

(2) When $$f(c) = 0$$ and $$f(a) = 2$$ then

$$3 \times 0 + 2 \times 2 + f(d) = f(b)$$

$$ \Rightarrow 4 + f(d) = f(b)$$

$$\therefore$$ possible value of $$f(d) = 1,3,4,5,6$$

$$\therefore$$ Total possible functions in this case = 5

(3) When $$f(c) = 0$$ and $$f(a) = 3$$ then

$$3 \times 0 + 2 \times 3 + f(d) = f(b)$$

$$ \Rightarrow 6 + f(d) = f(b)$$

$$\therefore$$ Possible value of $$f(d) = 1,2,4$$

$$\therefore$$ Total possible functions in this case = 3

(4) When $$f(c) = 0$$ and $$f(a) = 4$$ then

$$3 \times 0 + 2 \times 4 + f(d) = f(b)$$

$$ \Rightarrow 8 + f(d) = f(b)$$

$$\therefore$$ Possible value of $$f(d) = 1,2$$

$$\therefore$$ Total possible functions in this case = 2

(5) When $$f(c) = 0$$ and $$f(a) = 5$$ then

$$3 \times 0 + 2 \times 5 + f(d) = f(b)$$

$$ \Rightarrow 10 + f(d) = f(b)$$

Possible value of f(d) can be 0 but f(c) is already zero. So, no value to f(d) can satisfy.

$$\therefore$$ No function is possible in this case.

$$\therefore$$ 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

$$3 \times 1 + 2 \times 0 + f(d) = f(b)$$

$$ \Rightarrow 3 + f(d) = f(b)$$

$$\therefore$$ Possible value of $$f(d) = 2,3,4,5,6,7$$

$$\therefore$$ Total possible functions in this case = 6

(2) When $$f(c) = 1$$ and $$f(a) = 2$$ then

$$3 \times 1 + 2 \times 2 + f(d) = f(b)$$

$$ \Rightarrow 7 + f(d) = f(b)$$

$$\therefore$$ Possible value of $$f(d) = 0,3$$

$$\therefore$$ Total possible functions in this case = 2

(3) When $$f(c) = 1$$ and $$f(a) = 3$$ then

$$3 \times 1 + 2 \times 3 + f(d) = f(b)$$

$$ \Rightarrow 9 + f(d) = f(b)$$

$$\therefore$$ Possible value of $$f(d) = 0$$

$$\therefore$$ Total possible functions in this case = 1

$$\therefore$$ 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

$$3 \times 2 + 2 \times 0 + f(d) = f(b)$$

$$ \Rightarrow 6 + f(d) = f(b)$$

$$\therefore$$ Possible values of $$f(d) = 1,3,4$$

$$\therefore$$ Total possible functions in this case = 3

(2) When $$f(c) = 2$$ and $$f(a) = 1$$ then,

$$3 \times 2 + 2 \times 1 + f(d) = f(b)$$

$$ \Rightarrow 8 + f(d) = f(b)$$

$$\therefore$$ Possible values of $$f(d) = 0$$

$$\therefore$$ Total possible function in this case = 1

$$\therefore$$ 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

$$3 \times 3 + 2 \times 0 + f(d) = f(b)$$

$$ \Rightarrow 9 + f(d) = f(b)$$

$$\therefore$$ Possible values of $$f(d) = 1$$

$$\therefore$$ Total one-one functions from four cases

$$ = 17 + 9 + 4 + 1 = 31$$

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