$\beta={ }^7 C _4 \times 3 \times 2 \times{ }^5 C _2 \times 3!+{ }^7 C _3 \times 5 \times 4!$
$\therefore \frac{\alpha-\beta}{5!}=119.00$
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$I$. For any $n$, the roots are distinct.
$II$. There are infinitely many values of $n$ for which both roots are real.
$III$. The product of the roots is necessarily an integer.
$T_p=\left\{A=\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{a}\end{array}\right]: \mathrm{a}, \mathrm{b}, \mathrm{c} \in\{0,1, \ldots ., \mathrm{p}-1\}\right\}$
$1.$ The number of $A$ in $T_p$ such that $A$ is either symmetric or skew-symmetric or both, and $\operatorname{det}(\mathrm{A})$ divisible by $\mathrm{p}$ is
$(A)$ $(\mathrm{p}-1)^2$ $(B)$ $2(\mathrm{p}-1)$
$(C)$ $(\mathrm{p}-1)^2+1$ $(D)$ $2 \mathrm{p}-1$
$2.$ The number of $A$ in $T_p$ such that the trace of $A$ is not divisible by $p$ but det $(A)$ is divisible by $p$ is [Note: The trace of a matrix is the sum of its diagonal entries.]
$(A)$ $(\mathrm{p}-1)\left(\mathrm{p}^2-\mathrm{p}+1\right)$ $(B)$ $\mathrm{p}^3-(\mathrm{p}-1)^2$
$(C)$ $(\mathrm{p}-1)^2$ $(D)$ $(p-1)\left(p^2-2\right)$
$3.$ The number of $A$ in $T_p$ such that det $(A)$ is not divisible by $p$ is
$(A)$ $2 \mathrm{p}^2$ $(B)$ $p^3-5 p$ $(C)$ $p^3-3 p$ $(D)$ $p^3-p^2$
Give the answer question $1,2$ and $3.$