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Evaluate 20 C 5 + 5 ∑ R = 2 25 − R C 4 - Mathematics

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Question

Evaluate

\[^ {20}{}{C}_5 + \sum^5_{r = 2} {}^{25 - r} C_4\]

Solution

Given:
\[{}^{20} C_5 + \sum^5_{r = 2} 25 -^r C_4\]

\[{}^{20} C_5 + \sum^5_{r = 2} 25 -^r C_4 \]
\[ = {}^{20} C_5 +^{23} C_4 + {}^{22} C_4 + {}^{21} C_4 + {}^{20} C_4 \]
\[ = \left( {}^{20} C_{4_{}} + {}^{20} C_5 \right) +^{21} C_4 +^{22} C_4 +^{23} C_4 \]
\[=^{21} C_5 +^{21} C_4 + {}^{{}^{22}} C_4 +^{23} C_4\]

[∵\[{{}^{}}^n C_{r - 1} +^n C_r =^{n + 1} C_r\]]

\[= \left( {}^{21} C_{4_{}} + {}^{21} C_{5_{}} \right) +^{22} C_4 +^{23} C_4 \]

\[=^{22} C_5 +^{22} C_4 +^{23} C_4\]  [∵ \[{{}^{}}^n C_{r - 1} +^n C_r =^{n + 1} C_r\]]
\[=^{23} C_5 +^{22} C_4 +^{23} C_4\]
\[=^{23} C_5 +^{23} C_4 \]
\[ =^{24} C_5\]  [∵ \[{{}^{}}^n C_{r - 1} +^n C_r =^{n + 1} C_r\]]
\[= \frac{25!}{19! 5!} = \frac{24 \times 23 \times 22 \times 21 \times 20}{5 \times 4 \times 3 \times 2 \times 1} = 42504\]
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Factorial N (N!) Permutations and Combinations
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Chapter 17: Combinations - Exercise 17.1 [Page 8]

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RD Sharma Mathematics [English] Class 11
Chapter 17 Combinations
Exercise 17.1 | Q 19 | Page 8

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