The number of one-one functions f : {a, b, c, d} $\to$ {0, 1, 2, ......, 10} such

that 2f(a) $-$ f(b) + 3f(c) + f(d) = 0 is ___________.

<p>Given one-one function</p> <p>$f:\{ a,b,c,d\} \to \{ 0,1,2,\,\,....\,\,10\}$</p> <p>and $2f(a) - f(b) + 3f(c) + f(d) = 0$</p> <p>$\Rightarrow 3f(c) + 2f(a) + f(d) = f(b)$</p> <p>Case I:</p> <p>(1) Now let $f(c) = 0$ and $f(a) = 1$ then</p> <p>$3 \times 0 + 2 \times 1 + f(d) = f(b)$</p> <p>$\Rightarrow 2 + f(d) = f(b)$</p> <p>Now possible value of $f(d) = 2,3,4,5,6,7,$ and $8$.</p> <p>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.</p> <p>$\therefore$ Total possible functions when $f(c) = 0$ and $f(a) = 1$ are = 7</p> <p>(2) When $f(c) = 0$ and $f(a) = 2$ then</p> <p>$3 \times 0 + 2 \times 2 + f(d) = f(b)$</p> <p>$\Rightarrow 4 + f(d) = f(b)$</p> <p>$\therefore$ possible value of $f(d) = 1,3,4,5,6$</p> <p>$\therefore$ Total possible functions in this case = 5</p> <p>(3) When $f(c) = 0$ and $f(a) = 3$ then</p> <p>$3 \times 0 + 2 \times 3 + f(d) = f(b)$</p> <p>$\Rightarrow 6 + f(d) = f(b)$</p> <p>$\therefore$ Possible value of $f(d) = 1,2,4$</p> <p>$\therefore$ Total possible functions in this case = 3</p> <p>(4) When $f(c) = 0$ and $f(a) = 4$ then</p> <p>$3 \times 0 + 2 \times 4 + f(d) = f(b)$</p> <p>$\Rightarrow 8 + f(d) = f(b)$</p> <p>$\therefore$ Possible value of $f(d) = 1,2$</p> <p>$\therefore$ Total possible functions in this case = 2</p> <p>(5) When $f(c) = 0$ and $f(a) = 5$ then</p> <p>$3 \times 0 + 2 \times 5 + f(d) = f(b)$</p> <p>$\Rightarrow 10 + f(d) = f(b)$</p> <p>Possible value of f(d) can be 0 but f(c) is already zero. So, no value to f(d) can satisfy.</p> <p>$\therefore$ No function is possible in this case.</p> <p>$\therefore$ Total possible functions when $f(c) = 0$ and $f(a) = 1,2,3$ and $4$ are $= 7 + 5 + 3 + 2 = 17$</p> <p>Case II:</p> <p>(1) When $f(c) = 1$ and $f(a) = 0$ then</p> <p>$3 \times 1 + 2 \times 0 + f(d) = f(b)$</p> <p>$\Rightarrow 3 + f(d) = f(b)$</p> <p>$\therefore$ Possible value of $f(d) = 2,3,4,5,6,7$</p> <p>$\therefore$ Total possible functions in this case = 6</p> <p>(2) When $f(c) = 1$ and $f(a) = 2$ then</p> <p>$3 \times 1 + 2 \times 2 + f(d) = f(b)$</p> <p>$\Rightarrow 7 + f(d) = f(b)$</p> <p>$\therefore$ Possible value of $f(d) = 0,3$</p> <p>$\therefore$ Total possible functions in this case = 2</p> <p>(3) When $f(c) = 1$ and $f(a) = 3$ then</p> <p>$3 \times 1 + 2 \times 3 + f(d) = f(b)$</p> <p>$\Rightarrow 9 + f(d) = f(b)$</p> <p>$\therefore$ Possible value of $f(d) = 0$</p> <p>$\therefore$ Total possible functions in this case = 1</p> <p>$\therefore$ Total possible functions when $f(c) = 1$ and $f(a) = 0,2$ and $3$ are</p> <p>$= 6 + 2 + 1 = 9$</p> <p>Case III:</p> <p>(1) When $f(c) = 2$ and $f(a) = 0$ then</p> <p>$3 \times 2 + 2 \times 0 + f(d) = f(b)$</p> <p>$\Rightarrow 6 + f(d) = f(b)$</p> <p>$\therefore$ Possible values of $f(d) = 1,3,4$</p> <p>$\therefore$ Total possible functions in this case = 3</p> <p>(2) When $f(c) = 2$ and $f(a) = 1$ then,</p> <p>$3 \times 2 + 2 \times 1 + f(d) = f(b)$</p> <p>$\Rightarrow 8 + f(d) = f(b)$</p> <p>$\therefore$ Possible values of $f(d) = 0$</p> <p>$\therefore$ Total possible function in this case = 1</p> <p>$\therefore$ Total possible functions when $f(c) = 2$ and $f(a) = 0,1$ are</p> <p>$= 3 + 1 = 4$</p> <p>Case IV:</p> <p>(1) When $f(c) = 3$ and $f(a) = 0$ then</p> <p>$3 \times 3 + 2 \times 0 + f(d) = f(b)$</p> <p>$\Rightarrow 9 + f(d) = f(b)$</p> <p>$\therefore$ Possible values of $f(d) = 1$</p> <p>$\therefore$ Total one-one functions from four cases</p> <p>$= 17 + 9 + 4 + 1 = 31$</p>