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Permutations and Combinations (p-2) — Maths (commerce) STD 11 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 11 Commerce / ArtsMaths (commerce)Permutations and Combinations (p-2)2 Marks
Question
Find n if: ${ }^{2 n} C_3:{ }^n C_2=52: 3$
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Answer

$
\begin{array}{ll}
& { }^{2 n} \mathrm{C}_3:{ }^{\mathrm{n}} \mathrm{C}_2=52: 3 \\
\therefore \quad & \frac{(2 \mathrm{n}) !}{3 !(2 \mathrm{n}-3) !} \div \frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}=\frac{52}{3} \\
\therefore \quad & \frac{(2 \mathrm{n}) !}{3 !(2 \mathrm{n}-3) !} \times \frac{2 !(\mathrm{n}-2) !}{\mathrm{n} !}=\frac{52}{3} \\
\therefore \quad & \frac{(2 \mathrm{n})(2 \mathrm{n}-1)(2 \mathrm{n}-2)(2 \mathrm{n}-3) !}{3 \times 2 !(2 \mathrm{n}-3) !} \times \frac{2 !(\mathrm{n}-2) !}{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}=\frac{52}{3} \\
\therefore \quad & \frac{2 \mathrm{n}(2 \mathrm{n}-1) \cdot 2(\mathrm{n}-1)}{3} \times \frac{1}{\mathrm{n}(\mathrm{n}-1)}=\frac{52}{3} \\
\therefore & \frac{4(2 \mathrm{n}-1)}{3}=\frac{52}{3} \\
\therefore \quad & 2 \mathrm{n}-1=13 \\
\therefore \quad & 2 \mathrm{n}=14 \\
\therefore \quad & \mathrm{n}=7
\end{array}
$

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