Reader Alex Freuman passed this along — a simple method of establishing any row in Pascal’s triangle, attributed to Edric Cane. To establish, for example, the seventh row (after the initial solitary 1), create a row of fractions in which the numerators are 7, 6, 5, 4, 3, 2, 1 and the denominators are 1, 2, 3, 4, 5, 6, 7:

$\displaystyle \frac{7}{1} \times \frac{6}{2} \times \frac{5}{3} \times \frac{4}{4} \times \frac{3}{5} \times \frac{2}{6} \times \frac{1}{7}$

Now multiply these in sequence, cumulatively, to get the numbers for the seventh row of the triangle:

$\displaystyle 1 \quad 7 \quad 21 \quad 35 \quad 35 \quad 21 \quad 7 \quad 1$

These are the coefficients for

$\displaystyle \left ( a+b \right )^{7}=a^{7} + 7a^{6}b + 21a^{5}b^{2} + 35a^{4}b^{3} + 35a^{3}b^{4} + 21a^{2}b^{5} + 7ab^{6} + b^{7}$.

Cane writes, “It couldn’t be easier to remember or to implement.” Another example — row 10:

$\displaystyle \frac{10}{1} \times \frac{9}{2} \times \frac{8}{3} \times \frac{7}{4} \times \frac{6}{5} \times \frac{5}{6} \times \frac{4}{7} \times \frac{3}{8} \times \frac{2}{9} \times \frac{1}{10}$

$\displaystyle 1 \quad 10 \quad 45 \quad 120 \quad 210 \quad 252 \quad 210 \quad 120 \quad 45 \quad 10 \quad 1$