$\binom{n}{r} + \binom{n}{r-1}$ equals

1. This is PASCAL'S IDENTITY (a fundamental combinatorial identity): $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$. 2. Geometrically: it's the rule for building Pascal's triangle — each entry is the sum of the two entries directly above it. 3. Algebraic proof: $\binom{n}{r} = \dfrac{n!}{r!(n-r)!}$ and $\binom{n}{r-1} = \dfrac{n!}{(r-1)!(n-r+1)!}$. Common denominator and simplify yields $\dfrac{(n+1)!}{r!(n+1-r)!} = \binom{n+1}{r}$. 4. Combinatorial proof: number of ways to choose $r$ from $n+1$ objects = (ways INCLUDING object $n+1$) + (ways EXCLUDING it) = $\binom{n}{r-1} + \binom{n}{r}$. 5. Example: $n = 4, r = 2$: $\binom{4}{2} + \binom{4}{1} = 6 + 4 = 10 = \binom{5}{2}$ ✓. 6. Other options have wrong indices. _Source: NCERT Class 11 Mathematics, Ch 6, §6.4 (Pascal's identity), p. 11._