3. There are ten small balls of the same size, five of which are red and five are white. Now, these ten balls are arranged in a row arbitrarily, and numbered from left to right as $1,2, \cdots, 10$. Then the number of arrangements where the sum of the numbers of the red balls is greater than the sum of the numbers of the white balls is. $\qquad$
3. 126.

First, the sum of the numbers is $1+2+\cdots+10=$ 55. Therefore, the sum of the numbers on the red balls cannot be equal to the sum of the numbers on the white balls.

Second, if a certain arrangement makes the sum of the numbers on the red balls greater than the sum of the numbers on the white balls, then by swapping the positions of the red and white balls, we get an arrangement where the sum of the numbers on the red balls is less than the sum of the numbers on the white balls, and vice versa.

Therefore, the number of arrangements where the sum of the numbers on the red balls is greater than the sum of the numbers on the white balls is equal to the number of arrangements where the sum of the numbers on the red balls is less than the sum of the numbers on the white balls. Thus, the number of arrangements we are looking for is $\frac{1}{2} \mathrm{C}_{10}^{5}=126$.