Explanation
Total number of coins $$=6$$
(1) '0' heads $$=1$$ chance$$ \left( ^{ 6 }{ C }_{ 0 } \right) $$
(2) '1' heads $$=6$$ chance$$ \left( ^{ 6 }{ C }_{ 1 } \right) $$
(3) '2' heads$$ =15$$ chance$$ \left( ^{ 6 }{ C }_{ 2 } \right) $$
(4) '3' heads$$ =20$$ chance $$ \left( ^{ 6 }{ C }_{ 3 } \right) $$
Hence total outcome $$=42$$
$$ {\textbf{Step 1: Find m}} $$
$$ {\text{For m,}} $$
$$ {\text{First we select any 5 digits from 0,1,2,}}...{\text{,9}} $$
$$ {\text{Number of ways = }}{}^{10}{{\text{C}}_5} $$
$$ {\text{Now after selection there is only 1 way to arrange these selected digits, i}}{\text{.e}}{\text{., in descending order}}{\text{.}} $$
$$ {\text{Therefore m = }}{}^{10}{{\text{C}}_5}\times{\text{ 1 = }}{}^{10}{{\text{C}}_5} $$
$$ {\textbf{Step 2: Find n}} $$
$$ {\text{For n,First we select any 5 digits from 1,2,}}...{\text{,9}} $$
$$ {\text{We can't select zero as first digit because then the number won't be a 5 - digit number}}{\text{.}} $$
$$ {\text{Therefore number of ways = }}{}^9{{\text{C}}_5} $$
$$ \Rightarrow {\text{n = }}{}^9{{\text{C}}_5}\times{\text{ 1 = }}{}^9{{\text{C}}_5} $$
$$ \Rightarrow {\text{m - n = }}{}^{10}{{\text{C}}_5}{\text{ - }}{}^9{{\text{C}}_5} $$
$$ {\text{We know that,}}{}^n{{\text{C}}_r}{\text{ + }}{}^n{{\text{C}}_{r - 1}}{\text{ = }}{}^{n + 1}{{\text{C}}_r} $$
$$ \Rightarrow {}^{n + 1}{{\text{C}}_r}{\text{ - }}{}^n{{\text{C}}_r}{\text{ = }}{}^n{{\text{C}}_{r - 1}} $$
$$ \Rightarrow {}^{10}{{\text{C}}_5}{\text{ - }}{}^9{{\text{C}}_5}{\text{ = }}{}^9{{\text{C}}_4} $$
$$ {\text{Hence, m - n = }}{}^9{{\text{C}}_4} $$
$$ {\textbf{Hence, the correct answer is option A}} $$
Each place of a ten digit number can be fixed by any of the two digits. So, the number of ways to form a ten digit number is $${2^{10}}$$.
$$\begin{matrix} \sum _{ r=0 }^{ 10 }(r){ \, ^{ 20 } }{ C_{ r } } \\ \Rightarrow 0{ +^{ 20 } }{ C_{ 1 } }+{ 2^{ 20 } }{ C_{ 2 } }+{ 3^{ 20 } }{ C_{ 3 } }+............+{ 10^{ 20 } }{ C_{ 10 } } \\ \Rightarrow 20+20\times 19+\dfrac { { 20\times 19\times 18 } }{ 2 } +.........+10\times \dfrac { { 20! } }{ { 10!\times 10! } } \\ \Rightarrow 20\left[ { 1+19+\dfrac { { 19\times 18 } }{ 2 } +\dfrac { { 19\times 18\times 17 } }{ 6 } +...........+\dfrac { { 10\times 19! } }{ { 10!\times 10! } } } \right] \\ \Rightarrow 20\left[ { 1+19+\dfrac { { 19\times 18 } }{ 2 } +\dfrac { { 19\times 18\times 17 } }{ 6 } +.........+\dfrac { { 19! } }{ { 9!\times 10! } } } \right] \\ \, \, \, \, \, \, 20\left[ { ^{ 19 }{ C_{ 0 } }{ +^{ 19 } }{ C_{ 1 } }{ +^{ 19 } }{ C_{ 2 } }{ +^{ 19 } }{ C_{ 3 } }+..........{ +^{ 19 } }{ C_{ 10 } } } \right] \, \, \, \, Ans. \\ \end{matrix}$$
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