The number of one-one functions $$f : \{a, b, c, d\} \to \{0, 1, 2, \ldots, 10\}$$ such that $$2f(a) - f(b) + 3f(c) + f(d) = 0$$ is ______

Given,

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

Rearranging,

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

Since

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

is one-one, all values

$$f(a),f(b),f(c),f(d)$$

must be distinct.

Also,

$$f(b)\le10$$

Hence,

$$2f(a)+3f(c)+f(d)\le10$$

Now consider cases according to $$f(c).$$

Case 1: $$f(c)=0$$

Then,

$$f(b)=2f(a)+f(d)$$

Since values are distinct,

$$f(a),f(d)\neq0$$

Now count possible ordered pairs $$\big(f(a),f(d)\big).$$

For $$f(a)=1,$$

$$2+f(d)\le10\implies f(d)\le8$$

Excluding $$f(d)=1,$$

number of choices $$=7$$

For $$f(a)=2,$$

$$4+f(d)\le10\implies f(d)\le6$$

Excluding $$f(d)=2,$$

number of choices $$=5$$

For $$f(a)=3,$$

$$6+f(d)\le10\implies f(d)\le4$$

Excluding $$f(d)=3,$$

number of choices $$=3$$

For $$f(a)=4,$$

$$8+f(d)\le10\implies f(d)\le2$$

Number of choices $$=2$$

Total cases,

$$7+5+3+2=17$$

Case 2: $$f(c)=1$$

Then,

$$f(b)=2f(a)+f(d)+3$$

Now values cannot equal $$1.$$

For $$f(a)=0,$$

$$f(d)+3\le10\implies f(d)\le7$$

Excluding $$0,1,$$

choices $$=6$$

For $$f(a)=2,$$

$$4+f(d)+3\le10\implies f(d)\le3$$

Excluding $$1,2,$$

choices $$=2$$

For $$f(a)=3,$$

$$6+f(d)+3\le10\implies f(d)\le1$$

Excluding $$1,$$

choices $$=1$$

Total cases,

$$6+2+1=9$$

Case 3: $$f(c)=2$$

Then,

$$f(b)=2f(a)+f(d)+6$$

For $$f(a)=0,$$

$$f(d)+6\le10\implies f(d)\le4$$

Excluding $$0,2,$$

choices $$=3$$

For $$f(a)=1,$$

$$2+f(d)+6\le10\implies f(d)\le2$$

Excluding $$1,2,$$

choices $$=1$$

Total cases,

$$3+1=4$$

Case 4: $$f(c)=3$$

Then,

$$f(b)=2f(a)+f(d)+9$$

For $$f(a)=0,$$

$$f(d)+9\le10\implies f(d)\le1$$

Excluding $$0,$$

choices $$=1$$

Total cases $$=1$$

Therefore, total number of one-one functions is

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

Hence, the required number is

$$\boxed{31}$$.