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If 2nc3 : Nc2 = 44 : 3, Find N. - Mathematics

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If 2nC3 : nC2 = 44 : 3, find n.

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Solution

Given: 

\[2 n_{C_3} : n_{C_2} = 44: 3\]
\[\frac{2 n_{C_3}}{n_{C_2}} = \frac{44}{3}\]
\[ \Rightarrow \frac{2n!}{3! (2n - 3)!} \times \frac{2! (n - 2)!}{n!} = \frac{44}{3}\]
\[ \Rightarrow \frac{2n (2n - 1) (2n - 2)}{3 n (n - 1)} = \frac{44}{3}\]
\[ \Rightarrow (2n - 1) (2n - 2) = 22 (n - 1)\]
\[ \Rightarrow 4 n^2 - 6n + 2 = 22n - 22\]
\[ \Rightarrow 4 n^2 - 28n + 24 = 0\]
\[ \Rightarrow n^2 - 7n + 6 = 0\]
\[ \Rightarrow n^2 - 6n - n + 6 = 0\]
\[ \Rightarrow n (n - 6) - 1(n - 6) = 0\]
\[ \Rightarrow (n - 1) (n - 6) = 0\]
\[\Rightarrow n = 1\]   or,  
\[n = 6\]
Now, 
\[n = 1 \Rightarrow 2_{C_3} : 2_{C_2} = 44: 3\]
But, this is not possible.
∴ \[n = 6\]
Concept: Combination
  Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 17 Combinations
Exercise 17.1 | Q 13 | Page 8

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