What is the harmonic mean of the numbers C(10,3) , C(10,4) ,...

The given numbers are $^{10}C_{3}$, $^{10}C_{4}$, $^{10}C_{5}$, $^{10}C_{6}$, and $^{10}C_{7}$.

Calculating their values:
$^{10}C_{3} = \dfrac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$
$^{10}C_{4} = \dfrac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$
$^{10}C_{5} = \dfrac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$
$^{10}C_{6} = ^{10}C_{4} = 210$
$^{10}C_{7} = ^{10}C_{3} = 120$

The harmonic mean $H$ of $n$ numbers $x_1, x_2, \dots, x_n$ is given by:
$H = \dfrac{n}{\dfrac{1}{x_1} + \dfrac{1}{x_2} + \dots + \dfrac{1}{x_n}}$

Here $n = 5$. The sum of the reciprocals is:
$S = \dfrac{1}{120} + \dfrac{1}{210} + \dfrac{1}{252} + \dfrac{1}{210} + \dfrac{1}{120}$

$S = \dfrac{2}{120} + \dfrac{2}{210} + \dfrac{1}{252}$

$S = \dfrac{1}{60} + \dfrac{1}{105} + \dfrac{1}{252}$

Taking the LCM of $60, 105, 252$, which is $1260$:
$S = \dfrac{21 + 12 + 5}{1260} = \dfrac{38}{1260} = \dfrac{19}{630}$

The harmonic mean is:
$H = \dfrac{5}{S} = \dfrac{5}{\dfrac{19}{630}} = \dfrac{3150}{19}$

Answer: $3150/19$