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A student is allowed to choose atmost \(n\) books from a collection of \(2 n+1\) books. If the total number of ways in which he can select atleast one book is 255 , then the value of \(n\) is
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The correct answer is:
4
According to given information,
\(\begin{aligned}
& { }^{2 n+1} C_1+{ }^{2 n+1} C_2+{ }^{2 n+1} C_3+\ldots \ldots+{ }^{2 n+1} C_n=x \text{ Let}\\
& \because^{2 n+1} C_0+{ }^{2 n+1} C_1+{ }^{2 n+1} C_2+\ldots+{ }^{2 n+1} C_n+\ldots \\
& +{ }^{2 n+1} C_{2 n+1}=2^{2 n+1} \\
& \Rightarrow 2 x+{ }^{2 n+1} C_0+{ }^{2 n+1} C_{2 n+1}=2^{2 n+1} \\
& {\left[\because{ }^n C_{n-r}={ }^n C_r\right]} \\
& \Rightarrow \quad 2 x+1+1=2^{2 n+1} \\
& {\left[\because{ }^n C_0={ }^n C_n=1\right]} \\
& \Rightarrow \quad x=2^{2 n}-1=255 \quad \text{(given)}\\
& \Rightarrow \quad 2^{2 n}=256=2^8 \\
& \Rightarrow \quad n=4 \\
\end{aligned}\)
Hence, option (1) is correct.
\(\begin{aligned}
& { }^{2 n+1} C_1+{ }^{2 n+1} C_2+{ }^{2 n+1} C_3+\ldots \ldots+{ }^{2 n+1} C_n=x \text{ Let}\\
& \because^{2 n+1} C_0+{ }^{2 n+1} C_1+{ }^{2 n+1} C_2+\ldots+{ }^{2 n+1} C_n+\ldots \\
& +{ }^{2 n+1} C_{2 n+1}=2^{2 n+1} \\
& \Rightarrow 2 x+{ }^{2 n+1} C_0+{ }^{2 n+1} C_{2 n+1}=2^{2 n+1} \\
& {\left[\because{ }^n C_{n-r}={ }^n C_r\right]} \\
& \Rightarrow \quad 2 x+1+1=2^{2 n+1} \\
& {\left[\because{ }^n C_0={ }^n C_n=1\right]} \\
& \Rightarrow \quad x=2^{2 n}-1=255 \quad \text{(given)}\\
& \Rightarrow \quad 2^{2 n}=256=2^8 \\
& \Rightarrow \quad n=4 \\
\end{aligned}\)
Hence, option (1) is correct.
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