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If 16cr = 16cr + 2, Find Rc4.

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Question

If 16Cr = 16Cr + 2, find rC4.

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Solution

Given:

\[{}^{16} C_r = {}^{16} C_{r + 2}\]
\[16 = r + r + 2\]  [∵ Property 5: \[{}^n C_x = {}^n C_y \Rightarrow x = y\] or \[x + y = n\]
\[\Rightarrow 2r + 2 = 16\]
\[ \Rightarrow 2r = 14\]
\[ \Rightarrow r = 7\]
Now,
\[{}^r C_4 = {}^7 C_4\]
\[\Rightarrow {}^7 C_4 = {}^7 C_3\] [∵\[{}^n C_r =^n C_{n - r}\]
\[\Rightarrow^7 C_4 = {}^7 C_3 = \frac{7}{3} \times \frac{6}{2} \times \frac{5}{1} \times^4 C_0\]
[∵\[{}^n C_r = \frac{n}{r} . {}^{n - 1} C_{r - 1}\]]
\[\Rightarrow^7 C_4 = 35\] [∵\[{}^n C_0 = 1\]]
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Chapter 17: Combinations - Exercise 17.1 [Page 8]

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R.D. Sharma Mathematics [English] Class 11
Chapter 17 Combinations
Exercise 17.1 | Q 14 | Page 8

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