Smallest Number to appear 6 Times in Pascal's Triangle

Theorem
The smallest positive integer greater than $1$ to appear $6$ times in Pascal's Triangle is $120$.

Proof
We have:
 * $\dbinom {120} 1 = \dbinom {16} 2 = \dbinom {10} 3 = \dbinom {10} 7 = \dbinom {16} {14} = \dbinom {120} {119} = 120$

To verify that this is the smallest, we look at binomial coefficients that are no more than $120$.

Observe that for $n > 120$, $1 \le k \le n - 1$:
 * $\dbinom n k \ge \dbinom n 1 = n > 120$

For $n > 16$, $2 \le k \le n - 2$:
 * $\dbinom n k \ge \dbinom n 2 = \dfrac {n \paren {n - 1}} {2!} > \dfrac {16 \paren {15}} {2!} = 120$

For $n > 10$, $3 \le k \le n - 3$:
 * $\dbinom n k \ge \dbinom n 3 = \dfrac {n \paren {n - 1} \paren {n - 2}} {3!} > \dfrac {10 \paren 9 \paren 8} {3!} = 120$

Any number, except $2$, appears twice in the form $\dbinom n 1$ and $\dbinom n {n - 1}$.

Therefore we scout for numbers appearing four times, not as those forms, by listing $\dbinom n k$ for:
 * $n \le 10$ and $2 \le k \le n - 2$;
 * $n \le 16$ and $k = 2$ or $n - 2$

They are:
 * $6$
 * $10, 10$
 * $15, 20, 15$
 * $21, 35, 35, 21$
 * $28, 56, 70, 56, 28$
 * $36, 84, 126, 126, 84, 36$
 * $45, 120, *, *, *, 120, 45$
 * $55, 55$
 * $66, 66$
 * $78, 78$
 * $91, 91$
 * $105, 105$
 * $120, 120$

and we see that the only number appearing four times is $120$.