Explanation
$${\textbf{Step 1: Consider a matrix which has 4 elements.}}$$
$${\text{As a matrix contains four elements so, the order of a matrix can be,}}$$ $$1 \times 4$$ $${\text{or}}$$ $$4 \times 1$$ $${\text{or}}$$ $$2 \times 2.$$
$${\text{Also in every such matrix, each element is independent and has 4 different choices}}$$ $$\left( {0, 1, 2, 3} \right)$$
$$\therefore $$ $${\text{The number of ways to fill four places of a matrix of order 4}} \times {\text{1 by 0, 1, 2, 3}}$$
$$ = {}^4{C_1} \times {}^4{C_1} \times {}^4{C_1} \times {}^4{C_1}$$
$$\therefore $$ $${\text{The number of ways to fill four places of a matrix of order 4}} \times {\text{1 by 0, 1, 2, 3}}$$ $$ = 4 \times 4 \times 4 \times 4$$
$$\left(\mathbf {\because {}^n{C_1} = n} \right)$$
$$ = {4^4} \ldots \left( 1 \right)$$
$${\textbf{Step 2: Find total number of matrices that can be formed.}}$$
$${\text{From equation }}\left( 1 \right)$$ $${\text{we can say that, number of matrices in each order is }}{{\text{4}}^4}.$$
$${\text{Therefore,}}$$
$${\text{The number of matrices that can be formed}}$$ $$ = {4^4} + {4^4} + {4^4}$$
$$ \Rightarrow $$ $${\text{The number of matrices that can be formed}}$$ $$ = 3 \times {4^4}$$
$${\textbf{Hence, option (C)}}{\textbf{ 3}} \mathbf {\times {4^4}}$$ $${\textbf{is correct answer.}}$$
Paths are shown as :
Similarly if we start from A towards B we get another $$4$$ paths.
Similarly if we start from A towards B again $$3$$ paths.
∴ Total different paths $$4\times3=12$$
Exponent of $$7$$ in $$^{ 120 }{ C }_{ 50 }$$
$$^{ n }{ C }_{ r }=\dfrac { n! }{ (n-r)!r! } \\ \\ \Rightarrow ^{ 120 }{ C }_{ 50 }=\dfrac { 120! }{ 70!\times 50! } \\ =\dfrac { 120\times 119\times 118\times .....71 }{ 50\times 49\times 48\times .....2\times 1 } $$
Number of multiples of $$7$$ in Numerator $$=$$ Denominator $$=8$$
Therefore Exponent of $$7$$ is $$0$$
Please disable the adBlock and continue. Thank you.