Functions Part 1

Given one-one function$f: \{a, b, c, d\} \rightarrow \{0,1,2,\ldots,10\}$$\text{ and } 2 f(a)-f(b)+3 f(c)+f(d)=0$$\Rightarrow 3 f(c)+2 f(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$$=17+9+4+1=31$ Total one-one functions from four cases$=17+9+4+1=31$