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 $120$.

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 $120$.

$\blacksquare$