Daily math problems for September 26, 2026 — Combinatorics, Sequences, Probability & more

The binomial formula gives $(a+b{)}^2=a^2+2ab+b^2$.

$\bullet$ Expand the binomial coefficient using $C_n^k=\frac{n!}{k!(n-k)!}$:

$$C_{49}^6=\frac{49\cdot 48\cdot 47\cdot 46\cdot 45\cdot 44}{720}$$

$\bullet$ The product evaluates to:

$$C_{49}^6=13983816$$

$\bullet$ Take the reciprocal:

$$\frac{1}{C_{49}^6}=\frac{1}{13983816}$$

$S_{20}={\displaystyle \frac{20(2\cdot 2+19\cdot 3)}{2}}={\displaystyle \frac{20\cdot 61}{2}}=10\cdot 61=610$.

$P(A)=P(B)={\displaystyle \frac{1}{2}}$, $P(R\mathrm{\mid }A)={\displaystyle \frac{3}{4}}$, $P(R\mathrm{\mid }B)={\displaystyle \frac{1}{4}}$.

$P(A\mathrm{\mid }R)={\displaystyle \frac{{\displaystyle \frac{3}{4}}\cdot {\displaystyle \frac{1}{2}}}{{\displaystyle \frac{3}{4}}\cdot {\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{4}}\cdot {\displaystyle \frac{1}{2}}}}={\displaystyle \frac{3\mathrm{/}8}{4\mathrm{/}8}}={\displaystyle \frac{3}{4}}$.

By Viète's formulas, $r+s+t=6$ and $rs+rt+st=11$. Hence $r^2+s^2+t^2=36-22=14$.

Domain constraint: $x^2-4>0\Rightarrow x\in (-\mathrm{\infty },-2)\cup (2,+\mathrm{\infty })$.

Inequality: ${\log }_{1\mathrm{/}3}(x^2-4)>0={\log }_{1\mathrm{/}3}1$. Since the base ${\displaystyle \frac{1}{3}}<1$, ${\log }_{1\mathrm{/}3}$ is decreasing, so the inequality flips: $x^2-4<1$, i.e. $x^2<5$, so $x\in (-\sqrt{5},\sqrt{5})$.

Intersection with the domain: $[(-\mathrm{\infty },-2)\cup (2,+\mathrm{\infty })]\cap (-\sqrt{5},\sqrt{5})=(-\sqrt{5},-2)\cup (2,\sqrt{5})$.

$f(2)=2a+3=7\Rightarrow 2a=4\Rightarrow a=2$.

$\sigma (c\cdot x)=\mathrm{\mid }c\mathrm{\mid }\sigma (x)$. With $c=2$: $\sigma (2x)=2\sigma$.

$z^2=-4\Rightarrow z^2+4=0\Rightarrow (z-2i)(z+2i)=0$. Solutions: $z=2i$ or $z=-2i$.

$\det =(b-a)(c-a)(c-b)$. (It's zero iff at least two of $a,b,c$ are equal.)