Find a natural number greater than one that appears in Pascal's triangle

a) more than three times.

b) more than four times.

#
a) Any number of the form $m=C_{n}^{k}$, where $1<k<n / 2$, will satisfy this. It appears twice in the $n$-th row and twice in the $m$-th row. The smallest of these numbers is $10=C_{5}^{2}=C_{5}^{3}=C_{10}^{1}=C_{10}^{9}$.

b) First method. By examining the first few rows of Pascal's triangle, we notice that $120=C_{16}^{2}=C_{10}^{3}$. This number appears twice in the 10th, 16th, and 120th rows.

Second method. We will look for a number that can be represented both as $C_{n}^{2}=\frac{n(n-1)}{2}$ and as $C_{m+1}^{4}=\frac{(m+1) m(m-1)(m-2)}{4!}$. Then $12 n(n-1)=\left(m^{2}-1\right)\left(m^{2}-2 m\right)$. Suppose $3 n=m^{2}-2 m$ and $4(n-1)=m^{2}-1$. Subtracting, we get $n=2 m+3$. Substituting into the first equation, we get $m^{2}-8 m+9=0$, so $m=9, n=21$. Thus, $C_{21}^{2}=C_{10}^{4}=210$. Therefore, the number 210 appears twice in the 10th, 21st, and 210th rows.

Third method. We will look for three numbers of the form $m, 2 m, 3 m$ that appear consecutively in one row (then the number $3 m=m+2 m$ will also appear in the next row). If $m=C_{n}^{k}$, then $2 m=C_{n}^{k+1}=\frac{n-k}{k+1} \cdot m$, and $3 m=C_{n}^{k+2}=\frac{n-k-1}{k+2} \cdot 2 m$. Thus, $2 k+2=n-k, 3 k+6=2 n-2 k-2$. Solving the system, we find $k=4, n=14$. Therefore, the number $C_{n}^{k+2}=C_{14}^{6}=3003$ appears twice in the 14th, 15th (as $C_{15}^{5}$), and 3003rd rows.

## Answer

a) For example, 10, 15, 21. b) For example, 120, 210, 3003.