A Counting Problem

The Problem

Choose an integer $x$ from 0 to 999 randomly. Say the probability of the digits of $x$ containing 2 but no 3 is $\frac{\alpha }{1000}$. What is $\alpha$.

The Generalized Problem

We consider a generalized problem. Choose an integer $x$ from 0 to $10^n-1$ randomly. Say the probability of the digits of $x$ containing 2 but no 3 is $\frac{\alpha }{10^n}$. What is $\alpha$.

A Subtraction Solution

Let $N_1$ be the number of numbers whose digits don’t contain 3, $N_2$ be the number of numbers whose digits don’t contain 2 or 3. Then we know that

$$\alpha =N_1-N_2.$$

And we can see choosing a number randomly as choosing digits randomly. Therefore

$$\begin{array}{r}{N}_1=(10-1{)}^n,\\ N_2=(10-2{)}^n.\end{array}$$

Thus

$$\alpha =9^n-8^n.$$

When $n=3$, $\alpha =217$.

A Summation Solution

The previous solution is concise and the expression of $\alpha$ is elegant and scalable. However, even when $n$ is small, $9^n$ and $8^n$ are two large numbers that are hard to compute without the help of a calculator. Therefore we propose another solution that only involves the addition operation. Although the expression may be more complicated, the involved computations may be simpler when $n$ is small.

Let $N_k$ be the number of numbers whose digits contain $k$ 2s but no 3. Then we have

$$N_k=\left(\genfrac{}{}{0ex}{}{k}{n}\right)8^{n-k}.$$

Then

$$\alpha =\sum _{k=1}^n{N}_k=\sum _{k=1}^n\left(\genfrac{}{}{0ex}{}{k}{n}\right)8^{n-k}.$$

E.g., when $n=3$, the computation is

$$\alpha =3\times 8^2+3\times 8+1=217.$$

A MUCH SIMPLER Solution ¹

We use an array $(a,b,c)$ to denote three digits of a number without order. Then there are three cases:

$$\{\begin{array}{ll}(2,a,b),a\notin \{2,3\},b\notin \{2,3\}& :\left(\genfrac{}{}{0ex}{}{1}{3}\right)\times 8\times 8=192,\\ (2,2,a),a\notin \{2,3\}& :\left(\genfrac{}{}{0ex}{}{2}{3}\right)\times 8=24,\\ (2,2,2)& :1.\end{array}$$

So there are $(192+24+1)=217$ numbers satisfying the condition. Thus $\alpha =217$.

A Simple Solution

First we count the numbers whose digits contain 2. We use an array $(a,b,c)$ to denote three digits without order. There are three cases:

$$\{\begin{array}{ll}(2,a,b),a\ne 2,b\ne 2& :\left(\genfrac{}{}{0ex}{}{1}{3}\right)\times 9\times 9=243,\\ (2,2,a),a\ne 2& :\left(\genfrac{}{}{0ex}{}{2}{3}\right)\times 9=27,\\ (2,2,2)& :1.\end{array}$$

So there are total $(243+27+1)=271$ numbers whose digits contain 2.

Then we count the numbers whose digits contain both 2 and 3. There are still three cases

$$\{\begin{array}{ll}(2,3,a),a\ne 2,a\ne 3& :2\times \left(\genfrac{}{}{0ex}{}{2}{3}\right)\times 8=48,\\ (2,2,3)& :\left(\genfrac{}{}{0ex}{}{2}{3}\right)=3,\\ (2,3,3)& :\left(\genfrac{}{}{0ex}{}{1}{3}\right)=3.\end{array}$$

So there are total $(48+3+3)=54$ numbers whose digits contain both 2 and 3.

Therefore, there are $(271-54)=217$ numbers satisfying the condition. Thus $\alpha =217$.

Note that for illustration the solution is a little long. But the idea and all the operations are very simple that you don’t even need to write them down.

Footnotes

1. Compared to A Simple Solution ↩