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Consider the set $[n]=\{1,2,\dots,n\}$. Recall that the LHS counts the number of ways to choose $k$ elements from $[n]$, or in other words the number of $k$-sets of $[n]$. At the RHS the term $\binom{n-1}{k-1}$ counts the number of $k$-sets of $[n]$ that contain the element $n$ (note that $n$ "has been chosen" and we are only left to choose $k-1$ from the remaining $n-1$); the term $\binom{n-1}{k}$ counts the number of $k$-sets of $[n]$ that do not contain the element $n$ (all $k$ elements are chosen from $[n-1]$).

Below we illustrate the identity with an example. We set $n=6$ and $k=4$. The identity is now $\binom{6}{4}={\color{red}\binom{5}{3}}+{\color{blue}\binom{5}{4}}$ (or $15={\color{red}10}+{\color{blue}5}$ ). Sets that contain the element $6$ are colored red and sets that do not contain $6$ are colored blue.

0: {1, 2, 3, 4}

1: {1, 2, 3, 5}

2: {1, 2, 3, 6}

3: {1, 2, 4, 5}

4: {1, 2, 4, 6}

5: {1, 2, 5, 6}

6: {1, 3, 4, 5}

7: {1, 3, 4, 6}

8: {1, 3, 5, 6}

9: {1, 4, 5, 6}

10: {2, 3, 4, 5}

11: {2, 3, 4, 6}

12: {2, 3, 5, 6}

13: {2, 4, 5, 6}

14: {3, 4, 5, 6}

1: {1, 2, 3, 5}

2: {1, 2, 3, 6}

3: {1, 2, 4, 5}

4: {1, 2, 4, 6}

5: {1, 2, 5, 6}

6: {1, 3, 4, 5}

7: {1, 3, 4, 6}

8: {1, 3, 5, 6}

9: {1, 4, 5, 6}

10: {2, 3, 4, 5}

11: {2, 3, 4, 6}

12: {2, 3, 5, 6}

13: {2, 4, 5, 6}

14: {3, 4, 5, 6}