If $k_{1}$ and $k_{2}$ are $4$ digit numbers . find the number of ways of forming $k_{1}$ and $k_{2}$ so that $k_{2}$ can be
subtracted from $k_{1}$ without borrowing at any stage
could some help me with this, thanks
If $k_{1}$ and $k_{2}$ are $4$ digit numbers . number of ways of forming $k_{1}$ and $k_{2}$
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combinatorics
2 Answers
1
Every digit of $k_1$ should be greater than or equal to $k_2$.
Number of ways of choosing will be $(^{10}C_2+10)$.
The thousandth digit will be chosen from $\{1,..,9\}$ instead of $\{0,1,...9\}$.
No of ways to choose thousandth digit will be $(^9C_2+9)$
Hence total number of ways is
$$(^9C_2+9)\cdot(^{10}C_2+10)\cdot(^{10}C_2+10)\cdot(^{10}C_2+10)$$
1
We start by asking a simpler question: If $x$ and $y$ are chosen at random from $\{0,1,2, \dots ,9\}$, what is the probability that $x \le y$? To answer this, note that
$$\Pr(x Similarly, we can show that if $x$ and $y$ are chosen at random from $\{1,2,3, \dots ,9\}$ then $\Pr(x \le y) = 5/9$. So if we have two four-digit numbers $x_1 x_2 x_3 x_4$ and $y_1 y_2 y_3 y_4$ in decimal notation, the probability that $x_i \le y_i$ for $i=1,2,3,4$ is
$$\frac{5}{9} \times \left( \frac{11}{20} \right)^3 \approx 0.09243$$