Show that `|^x C_r^x C_(r+1)^x C_(r+2)^y C_r^y C_(r+1)^y C_(r+2)^z C_r^z C_(r+1)^z C_(r+2)|=|^x C_r^(x+1)C_(r+1)^(x+2)C_(r+2)^y C_r^(y+1)C_(r+1)^(y+2)C_(r+2)^z C_r^(z+1)C_(r+1)^(z+2)C_(r+2)|` .

Options

A. `0`
B. `2^(n)`
C. `"^(x+y+z)C_(r )`
D. `"^(x+y+z)C_(r +2)`

Answer

A

Explanation

`(a)` `|{:(.^(x)C_(r),.^(x)C_(r+1),.^(x)C_(r+2)),(.^(y)C_(r ),.^(y)C_(r+1),.^(y)C_(r+2)),(.^(z)C_(r ),.^(z)C_(r+1),.^(z)C_(r+2)):}|-|{:(.^(x)C_(r),.^(x+1)C_(r+1),.^(x+2)C_(r+2)),(.^(y)C_(r ),.^(y+1)C_(r+1),.^(y+2)C_(r+2)),(.^(z)C_(r ),.^(z+1)C_(r+1),.^(z+2)C_(r+2)):}|`
In first determinant apply `C_(3)toC_(3)+C_(2)` and `C_(2)toC_(2)+C_(1)` and then agan `C_(3)toC_(3)+C_(2)`