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If \(1 \times 1 !+2 \times 2 !+3 \times 3 !+\ldots+n \times n !=11 !-1\), then the maximum value of \({ }^n C_r\) is
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The correct answer is:
252
\(\begin{aligned}
& 1 \times 1 !+2 \times 2 !+\ldots+n \times n ! \\
& =(2-1) \times 1 !+(3-1) \times 2 !+\ldots+[(n+1)-1] \times n ! \\
& =\{2 \times 1 !+3 \times 2 !+4 \times 3 !+\ldots(n+1) \times n !\} \\
& \quad-\{1 \times 1 !+1 \times 2 !+1 \times 3 !+\ldots+1 \times n !\} \\
& =(n+1) !-1 !=11 !-1 ! \quad \quad \text { (given) }
\end{aligned}\)
So, \(n=10\)
Now, maximum value of \({ }^{10} C_r\) occurs when
\(r=\frac{n}{2}=\frac{10}{2}=5\)
Now, \({ }^{10} C_5=\frac{10 !}{5 !(10-5) !}=252\)
& 1 \times 1 !+2 \times 2 !+\ldots+n \times n ! \\
& =(2-1) \times 1 !+(3-1) \times 2 !+\ldots+[(n+1)-1] \times n ! \\
& =\{2 \times 1 !+3 \times 2 !+4 \times 3 !+\ldots(n+1) \times n !\} \\
& \quad-\{1 \times 1 !+1 \times 2 !+1 \times 3 !+\ldots+1 \times n !\} \\
& =(n+1) !-1 !=11 !-1 ! \quad \quad \text { (given) }
\end{aligned}\)
So, \(n=10\)
Now, maximum value of \({ }^{10} C_r\) occurs when
\(r=\frac{n}{2}=\frac{10}{2}=5\)
Now, \({ }^{10} C_5=\frac{10 !}{5 !(10-5) !}=252\)
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