Passcodes and Repeated Digits

The content in this post is taken from this video on the YouTube channel MindYourDecisions. Figure 1 shows if the digits of a four digit code are known, there are 24 possibilites of the code contains four distinct digits but 36 possibilities if the code contains only three digits (with one digit repeated).

Figure 1

If no digits are repeated, then there are:$$4! = 4 \times 3 \times 2 \times 1 =24 \text{ possibilities}$$However if one digit is repeated then we have:$$ \frac{4!}{2!} \times 3 = 36 \text{ possibilities}$$What happens with a six digit passcode when we know the digits but not the order. If six digits are used, then we have:$$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1  = 720 \text{ possibilities}$$However, if one digit is repeated then we have:$$ \frac{6!}{2!} \times 5 =1800 \text{ possibilities: see Figure 2}$$

Figure 2

The formulae in general for distinct digits versus one repeated digit are:$$ n! \leftarrow \text{ versus } \rightarrow \frac{(n-1) \times n!}{2}$$Once we have \(n \geq 4\) then \( (n-1)/2 \) is greater than 1 and so the single repeated digit passcode will always yield more possibilities than the distinct digits. What about repeating more than one digit? 

Figure 3 shows the situation for six digit passcodes where it can be seen that one repeated digit yields the most possibilities:

Figure 3

However, for a seven digit posscode, using five distinct digits (with two digits repeated) yields the most possibilities. See Figure 4.

Figure 4

The video referenced earlier concludes with a mention of the general case and a rather formidable formula that I won't go into here.