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If Nc10 = Nc12, Find 23cn. - Mathematics

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If nC10 = nC12, find 23Cn.

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Solution

Given: nC10 = nC12
We have,

\[{}^n C_{10} = {}^n C_{12}\]
\[\Rightarrow n = 12 + 10 = 22\]  [∵\[{}^n C_x = {}^n C_y \Rightarrow x = y\]]  or, 
\[n = x + y\]
Now,
\[{}^{23} C_{22} = {}^{23} C_1\] [∵ \[{}^n C_r = {}^n C_{n - r}\]]
\[\Rightarrow {}^{23} C_{22} =^{23} C_1 = \frac{23}{1} \times^{22} C_0\] [∵\[{}^n C_r = \frac{n}{r} {}^{n - 1} C_{r - 1}\]]
\[\Rightarrow^{23} C_{22} = 23\] [∵\[{}^n C_0 = 1\]]
Concept: Combination
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 17 Combinations
Exercise 17.1 | Q 4 | Page 8

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