The question: How many $10$ character codes you can create using a-z, A-Z, 0-9 and 10 symbols (!@#$%...) with at least one symbol?

The correct answer: $72^{10}-62^{10}$, counting all possible combinations minus the combinations without any symbol

My way of thinking: I put to the first place a symbol ($10$ ways), for the rest 9 positions a put whatever I want ($72^9$), I multiply by 10 to get every possible position of the symbol. $10^2 \cdot 72^9$ What's wrong in my thinking? What am I double counting? Is there a way to find the same solution using my thinking?

  • 2
    $\begingroup$ Consider 11111111!!, which of the !'s is your "first place"? $\endgroup$ – peterwhy Aug 9 '17 at 10:01
  • $\begingroup$ that's why i multiplied by 10 to get every possible position of the symbol. $\endgroup$ – user285936 Aug 9 '17 at 10:02
  • $\begingroup$ My point was there are two !'s, and either can be your "first place", which is the source of double counting. $\endgroup$ – peterwhy Aug 9 '17 at 10:04
  • $\begingroup$ There is an alternative route to find the same solution: for $i=1,\dots,10$ find the number of codes having exactly $i$ symbols, and then take the summation. Not recommended though. $\endgroup$ – drhab Aug 9 '17 at 10:05
  • $\begingroup$ @drhab Σ(62^10-n * 10^n ) $\endgroup$ – user285936 Aug 9 '17 at 10:11

A slight modification to your thinking:

Give up the "multiply by 10 to get every possible position of the symbol", but say the first symbol appears on position $i$, $1\le i \le 10$.

For each position $i$, each character before it has $62$ independent choices, and each character after it has $72$ independent choices. The number of choices is

$$\begin{align*} N &= \sum_{i = 1}^{10} 62^{i-1}\cdot 10 \cdot 72^{10-i}\\ &= \sum_{j = 0}^{9} 62^{j}\cdot 10 \cdot 72^{9-j}\\ &= 10\cdot72^9\sum_{j=0}^{9}\left(\frac{62}{72}\right)^j\\ &= 10\cdot72^9\cdot\frac{1-(62/72)^{10}}{1-62/72}\\ &= 10\cdot \frac{72^{10}-62^{10}}{72-62}\\ &= 72^{10}-62^{10} \end{align*}$$

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  • $\begingroup$ Ok, now it is clearer. thank you everyone $\endgroup$ – user285936 Aug 9 '17 at 10:21

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