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Let R and N Be Positive Integers Such that 1 ≤ R ≤ N. Then Prove the Following: Ncr + 2 · Ncr − 1 + Ncr − 2 = N + 2cr. - Mathematics

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Question

Let r and n be positive integers such that 1 ≤ r ≤ n. Then prove the following:

 nCr + 2 · nCr − 1 + nCr − 2 = n + 2Cr.

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Solution

\[{}^n C_r + 2 . {}^n C_{r - 1} +^n C_{r - 2} = {}^{n + 2} C_r\]

\[LHS =^n C_r + 2 .^n C_{r - 1} +^n C_{r - 2} \]
\[ = \left( {}^n C_r +^n C_{r - 1} \right) + \left( {}^n C_{r - 1} +^n C_{r - 2} \right)\]

\[= {}^{n + 1} C_r + {}^{n + 1} C_{r - 1}\] [∵ \[n_{C_r} + n_{C_{r - 1}} = n + 1_{C_r}\]]
\[=^{n + 2} C_r\] [∵\[n_{C_r} + n_{C_{r - 1}} = n + 1_{C_r}\]]
 = RHS
∴  LHS = RHS
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Factorial N (N!) Permutations and Combinations
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Chapter 17: Combinations - Exercise 17.1 [Page 9]

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RD Sharma Mathematics [English] Class 11
Chapter 17 Combinations
Exercise 17.1 | Q 20.4 | Page 9

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