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Given the digits $0,1,...,9$, the numer of possibilities to construct a PIN code is $10^4$ (zero can be at the beginning since it's a code and not a number).

I need to find the number of PIN codes that have at least two of the same digits.

I tried using two ways to calculate, but got different results and I could really use some help figuring out why it happend.

First solution:

The number of PIN codes with all distinct digits is $10\times9\times8\times7$. So the number of PIN codes with at least two of the same digits should be $10^4 - 10\times9\times8\times7 = 4960$.

Second solution:

If a PIN code has at least two of the same digits, it has exactly two, three or four of the same digits.

For having exactly two of the same digits, I have $10$ choices for what digits should be the same. I then choose their position in the code, ${4 \choose 2}$ choices for that , and then there are $9\times8$ choices for the other two digits. All in all, there are $10\times{4 \choose 2}\times8\times9=4320$ possibilities for that.

For having exactly three of the same digits, again there are $10$ choices for what digits should be the same. For choosing their position in the code there are ${4 \choose 3}$ ways, and $9$ ways to pick the fourth digit. So there are $10\times{4 \choose 3}\times9 = 360$ possibilities for that.

Lastly, the number of possible ways to construct a PIN code with exactly 4 of the same digits is $10$, which is trivial $(0000,1111,...,9999)$.

Summing up all of the above, we get that the number of possibilities is $4320+360+10 = 4690$.

I could really use some help figuring out why one (or both) of my methods is wrong.

Thanks!

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  • $\begingroup$ Doesn't $0000$ have only one distinct digit? You have totally misinterpreted the question. $\endgroup$
    – Parcly Taxel
    Jan 6 at 16:15
  • $\begingroup$ Hi, my apologies, since English is not my main language I totally misinterpreted the meaning of the word distinct. I edited it, hope it makes sence now. $\endgroup$
    – Orokusaki
    Jan 6 at 16:23
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    $\begingroup$ You missed the case where you have two pairs, for example 1122 or 3443. $\endgroup$
    – Jaap Scherphuis
    Jan 6 at 16:26
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    $\begingroup$ No pairs give you $270$ and not $540$. You are double counting. $\endgroup$
    – Math Lover
    Jan 6 at 17:14
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    $\begingroup$ Say you choose $1$ and the next number is $2$. Your $4C2$ will count 1 2 1 2, 2 1 2 1 and so on. Now when the first number is $2$ and the next number is $1$, you again count them. $\endgroup$
    – Math Lover
    Jan 6 at 17:19
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Ok, so thanks for the help guys, it turns out I wasn't counting pairs of the same digit. The way I calculated it:

We have $10\times9$ choices for the two pairs of identical digits. Next, we need to find out how many possibilities are there to divide them to pairs. This is much like dividng $2n$ people to pairs, when $n=2$.

All in all we get: $10\times9\times\frac{4!}{2^{2}\times2!} = 270$.

Summing this with the above we get $270+4690=4960$, which is the correct answer.

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