Computing Binomial Coefficients

Worked examples for r = ±1/2, ±1/3, ±1/4, ±3/2

Published

November 10, 2026

For any real number \(r\), the binomial coefficient is

\[ \binom{r}{k} = \frac{r(r-1)(r-2)\cdots(r-k+1)}{k!}, \qquad \binom{r}{0} = 1. \]

There are \(k\) factors on top, starting at \(r\) and going down by \(1\) each time.

The method for fractions

Write every factor over the common denominator \(q\), pull out \(q^k\), then count the minus signs.

\(r = \tfrac12\)

Factors: \(\tfrac12,\ -\tfrac12,\ -\tfrac32,\ -\tfrac52,\ \ldots\)

\(\binom{1/2}{1} = \tfrac12\)
\(\binom{1/2}{2} = \dfrac{(\frac12)(-\frac12)}{2!} = \dfrac{-1/4}{2} = -\dfrac18\)
\(\binom{1/2}{3} = \dfrac{(\frac12)(-\frac12)(-\frac32)}{3!} = \dfrac{3/8}{6} = \dfrac1{16}\)
\(\binom{1/2}{4} = \dfrac{(\frac12)(-\frac12)(-\frac32)(-\frac52)}{4!} = \dfrac{-15/16}{24} = -\dfrac{5}{128}\)

Pattern (\(k \ge 2\)): \[ \binom{1/2}{k} = \frac{(-1)^{k+1}\cdot 1\cdot3\cdot5\cdots(2k-3)}{2^k\,k!} \]

Factors: \(-\tfrac12,\ -\tfrac32,\ -\tfrac52,\ -\tfrac72,\ \ldots\) Every factor is negative.

\(\binom{-1/2}{1} = -\tfrac12\)
\(\binom{-1/2}{2} = \dfrac{(-\frac12)(-\frac32)}{2} = \dfrac{3/4}{2} = \dfrac38\)
\(\binom{-1/2}{3} = \dfrac{(-\frac12)(-\frac32)(-\frac52)}{6} = \dfrac{-15/8}{6} = -\dfrac5{16}\)
\(\binom{-1/2}{4} = \dfrac{105/16}{24} = \dfrac{35}{128}\)

Pattern: \[ \binom{-1/2}{k} = \frac{(-1)^k\cdot 1\cdot3\cdot5\cdots(2k-1)}{2^k\,k!} \]

Factors: \(\tfrac13,\ -\tfrac23,\ -\tfrac53,\ -\tfrac83,\ \ldots\)

\(\binom{1/3}{1} = \tfrac13\)
\(\binom{1/3}{2} = \dfrac{(\frac13)(-\frac23)}{2} = \dfrac{-2/9}{2} = -\dfrac19\)
\(\binom{1/3}{3} = \dfrac{(\frac13)(-\frac23)(-\frac53)}{6} = \dfrac{10/27}{6} = \dfrac5{81}\)
\(\binom{1/3}{4} = \dfrac{-80/81}{24} = -\dfrac{10}{243}\)

Pattern (\(k \ge 2\)): \[ \binom{1/3}{k} = \frac{(-1)^{k+1}\cdot 1\cdot2\cdot5\cdots(3k-4)}{3^k\,k!} \]

Factors: \(-\tfrac13,\ -\tfrac43,\ -\tfrac73,\ -\tfrac{10}3,\ \ldots\)

\(\binom{-1/3}{1} = -\tfrac13\)
\(\binom{-1/3}{2} = \dfrac{(-\frac13)(-\frac43)}{2} = \dfrac{4/9}{2} = \dfrac29\)
\(\binom{-1/3}{3} = \dfrac{(-\frac13)(-\frac43)(-\frac73)}{6} = \dfrac{-28/27}{6} = -\dfrac{14}{81}\)
\(\binom{-1/3}{4} = \dfrac{280/81}{24} = \dfrac{35}{243}\)

Pattern: \[ \binom{-1/3}{k} = \frac{(-1)^k\cdot 1\cdot4\cdot7\cdots(3k-2)}{3^k\,k!} \]

Factors: \(\tfrac14,\ -\tfrac34,\ -\tfrac74,\ -\tfrac{11}4,\ \ldots\)

\(\binom{1/4}{1} = \tfrac14\)
\(\binom{1/4}{2} = \dfrac{(\frac14)(-\frac34)}{2} = \dfrac{-3/16}{2} = -\dfrac3{32}\)
\(\binom{1/4}{3} = \dfrac{(\frac14)(-\frac34)(-\frac74)}{6} = \dfrac{21/64}{6} = \dfrac7{128}\)
\(\binom{1/4}{4} = \dfrac{-231/256}{24} = -\dfrac{77}{2048}\)

Pattern (\(k \ge 2\)): \[ \binom{1/4}{k} = \frac{(-1)^{k+1}\cdot 1\cdot3\cdot7\cdots(4k-5)}{4^k\,k!} \]

Factors: \(-\tfrac14,\ -\tfrac54,\ -\tfrac94,\ -\tfrac{13}4,\ \ldots\)

\(\binom{-1/4}{1} = -\tfrac14\)
\(\binom{-1/4}{2} = \dfrac{(-\frac14)(-\frac54)}{2} = \dfrac{5/16}{2} = \dfrac5{32}\)
\(\binom{-1/4}{3} = \dfrac{(-\frac14)(-\frac54)(-\frac94)}{6} = \dfrac{-45/64}{6} = -\dfrac{15}{128}\)
\(\binom{-1/4}{4} = \dfrac{585/256}{24} = \dfrac{195}{2048}\)

Pattern: \[ \binom{-1/4}{k} = \frac{(-1)^k\cdot 1\cdot5\cdot9\cdots(4k-3)}{4^k\,k!} \]

Factors: \(\tfrac32,\ \tfrac12,\ -\tfrac12,\ -\tfrac32,\ \ldots\) There are two positive factors before the signs flip.

\(\binom{3/2}{1} = \tfrac32\)
\(\binom{3/2}{2} = \dfrac{(\frac32)(\frac12)}{2} = \dfrac{3/4}{2} = \dfrac38\)
\(\binom{3/2}{3} = \dfrac{(\frac32)(\frac12)(-\frac12)}{6} = \dfrac{-3/8}{6} = -\dfrac1{16}\)
\(\binom{3/2}{4} = \dfrac{(\frac32)(\frac12)(-\frac12)(-\frac32)}{24} = \dfrac{9/16}{24} = \dfrac3{128}\)

Pattern (\(k \ge 3\)): \[ \binom{3/2}{k} = \frac{(-1)^k\cdot 3\cdot 1\cdot\big(1\cdot3\cdots(2k-5)\big)}{2^k\,k!} \]

Factors: \(-\tfrac32,\ -\tfrac52,\ -\tfrac72,\ -\tfrac92,\ \ldots\)

\(\binom{-3/2}{1} = -\tfrac32\)
\(\binom{-3/2}{2} = \dfrac{(-\frac32)(-\frac52)}{2} = \dfrac{15/4}{2} = \dfrac{15}8\)
\(\binom{-3/2}{3} = \dfrac{(-\frac32)(-\frac52)(-\frac72)}{6} = \dfrac{-105/8}{6} = -\dfrac{35}{16}\)
\(\binom{-3/2}{4} = \dfrac{945/16}{24} = \dfrac{315}{128}\)

Pattern: \[ \binom{-3/2}{k} = \frac{(-1)^k\cdot 3\cdot5\cdot7\cdots(2k+1)}{2^k\,k!} \]

\(r\)	\(k = 1\)	\(k = 2\)	\(k = 3\)	\(k = 4\)
\(\tfrac12\)	\(\tfrac12\)	\(-\tfrac18\)	\(\tfrac1{16}\)	\(-\tfrac5{128}\)
\(-\tfrac12\)	\(-\tfrac12\)	\(\tfrac38\)	\(-\tfrac5{16}\)	\(\tfrac{35}{128}\)
\(\tfrac13\)	\(\tfrac13\)	\(-\tfrac19\)	\(\tfrac5{81}\)	\(-\tfrac{10}{243}\)
\(-\tfrac13\)	\(-\tfrac13\)	\(\tfrac29\)	\(-\tfrac{14}{81}\)	\(\tfrac{35}{243}\)
\(\tfrac14\)	\(\tfrac14\)	\(-\tfrac3{32}\)	\(\tfrac7{128}\)	\(-\tfrac{77}{2048}\)
\(-\tfrac14\)	\(-\tfrac14\)	\(\tfrac5{32}\)	\(-\tfrac{15}{128}\)	\(\tfrac{195}{2048}\)
\(\tfrac32\)	\(\tfrac32\)	\(\tfrac38\)	\(-\tfrac1{16}\)	\(\tfrac3{128}\)
\(-\tfrac32\)	\(-\tfrac32\)	\(\tfrac{15}8\)	\(-\tfrac{35}{16}\)	\(\tfrac{315}{128}\)

Takeaways

Negative \(r\): every factor is negative, so the signs simply alternate as \((-1)^k\), and the product has no “missing” factor.
Positive \(r\) between \(0\) and \(1\): the first factor is positive and the rest are negative, so the sign is \((-1)^{k+1}\) for \(k \ge 1\). The last factor in the product is one step smaller than you’d guess: \((2k-3)\), \((3k-4)\), \((4k-5)\).
\(r = \tfrac32\): two factors are positive before the signs start alternating, so the general pattern only kicks in at \(k = 3\).
A quick check: always compute \(\binom r2\) by hand and compare it with whatever general formula you write down.