Daily math problems for June 22, 2026 — Combinatorics, Algebra

$\left(\genfrac{}{}{0px}{}{10}{3}\right)={\displaystyle \frac{10\cdot 9\cdot 8}{3!}}={\displaystyle \frac{720}{6}}=120$.

$\bullet$ Take reciprocals of each equation, using $\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$:

$$\frac{1}{x}+\frac{1}{y}=\frac{3}{5},\frac{1}{x}+\frac{1}{z}=\frac{2}{3},\frac{1}{y}+\frac{1}{z}=\frac{1}{4}$$

$\bullet$ Let $a=1\mathrm{/}x$, $b=1\mathrm{/}y$, $c=1\mathrm{/}z$. Sum all three:

$$2(a+b+c)=\frac{3}{5}+\frac{2}{3}+\frac{1}{4}=\frac{91}{60}$$

$$a+b+c=\frac{91}{120}$$

$\bullet$ Subtract each pair-sum from $a+b+c$:

$$a=\frac{61}{120},b=\frac{11}{120},c=\frac{19}{120}$$

$\bullet$ Recover $x,y,z$ and add:

$$x+y+z=\frac{120}{61}+\frac{120}{11}+\frac{120}{19}$$

$S_3$ (or equivalently $D_3$) has $6$ elements and is non-abelian. All groups of order $1,2,3,4,5$ are abelian.