$$ \text { Match the items of List-I to the items of List-II } $$
| List-I | List-II | ||
|---|---|---|---|
| (A) | The number of ways of not selecting ( $n-r$ ) things from $n$ different things | (I) | $1+{ }^n C_1+{ }^n C_2+\ldots+{ }^n C_r$ |
| (B) | $\quad(n-r+1) \cdot{ }^n C_{r-1}$ | (II) | $(r+1) \cdot{ }^n C_{r+1}$ |
| (C) | The number of ways of selecting atleast ( $n-r$ ) things from $n$ different things | (III) | $r \cdot{ }^n \mathrm{C}$, |
| (D) | $(n-r)\left({ }^{(n-1)} C_{r-1}+{ }^{(n-1)} C_r\right)$ | (IV) | $$ \begin{aligned} & 2^n-1-n- \\ & { }^n C_2-\ldots-{ }^n C_r \end{aligned} $$ |
| (V) | ${ }^n C_{n-1}$ | ||
Consider the following statements:
I. The number of positive integral solutions of $x_1+x_2+x_3+x_4=10$ is 286 .
II. If $25!=10^n \times k,(k \in \mathbf{N})$, then $n=6$
Which one of the following options is true?
A student is allowed to select at least $(n+1)$ books but not all books from a collection of ( $2 n+1$ ) books. If the total number of ways in which he can select these books is 255 , then the number of books in that collection is
Let $S_r=\{x, y, z) / x+y+z=11, x \geq r, y \geq r$, $z \geq r, x, y, z, r$ are integers $\}$ and $n\left(S_r\right)$ represents the number of elements in $S_r$. Then $n\left(S_{2)}+n\left(S_3\right)+n\left(S_4\right)=\right.$
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