Prove that ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$

**OR**

(i) How many permutations can be made of the letters of the word "SERIES"? (ii) How many of these start with S and end with S?

Question

Prove that ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$

**OR**

(i) How many permutations can be made of the letters of the word "SERIES"? (ii) How many of these start with S and end with S?

Accepted Answer

>

> [!TLDR]
> Combining the two combination terms algebraically over a common factor reduces exactly to $^{n+1}C_r$ — this is Pascal's identity.
>
> We start from the definitions:
> $$^{n}C_r=\frac{n!}{r!(n-r)!},\qquad {}^{n}C_{r-1}=\frac{n!}{(r-1)!(n-r+1)!}$$
>
> Add them, factoring out the common part $\dfrac{n!}{(r-1)!(n-r)!}$:
> $$^{n}C_r+{}^{n}C_{r-1}=\frac{n!}{(r-1)!(n-r)!}\left[\frac{1}{r}+\frac{1}{n-r+1}\right]$$
>
> Combine the bracketed fraction over a common denominator $r(n-r+1)$:
> $$\frac{1}{r}+\frac{1}{n-r+1}=\frac{(n-r+1)+r}{r(n-r+1)}=\frac{n+1}{r(n-r+1)}$$
>
> So: …

Q.Prove that ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$ OR

Concept understanding — Combinations Symmetry Property

The Intuition: Two Ways to Choose

The Precise Statement

Why It Works (The Algebraic Proof)

Special Cases That Make Sense

Q.Prove that nCr+nCr−1=n+1Cr{}^nC_r + {}^nC_{r-1} = {}^{n+1}C_rnCr​+nCr−1​=n+1Cr​ OR

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Q.Prove that ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$ OR

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