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Suppose a standard combination lock on a door has 5 distinct buttons (labelled e.g. 1, 2, 3, 4, 5).

A passcode is defined by 5 button presses.
Buttons can be repeated, so e.g. (5,5,5,5,5), (4,4,4,5,4), (1,2,4,3,5) are all valid passcodes.

The deluxe version of the lock allows "joins". A join is when two distinct buttons are pressed at the same time, so for example the four following codes are all valid and distinct:

  • (1, 2, 3, 4+5),
  • (1, 2+3, 4+5),
  • (1, 2+3, 4, 5), and
  • (1, 2, 3, 4, 5).

How many passcodes are possible in the standard and deluxe locks?

My attempt:

Standard lock: You can choose any character 5 times, so there are $5^5 = 3125$ standard passcodes.

Deluxe lock: Deluxe lock codes have either 0, 1, or 2 joins.

The 0-join codes are the 3125 for the standard lock.

The 1-join codes can have a join in one of four positions (between characters 1&2, 2&3, 3&4, 4&5). Having one join only constrains one of the characters (it has to be different to the character preceeding it). So for each of the four 1-join-positions, there are $5^4 \times 4$ choices of codes, so there are $5^4 \times 4^2 = 10000$ 1-join codes.

The 2-join codes must either be between characters 1&2 and 3&4, or 1&2 and 4&5, or 2&3 and 3&4. Having 2 joins constrains each of two of the characters to be different to the character preceeding it. So for each of the three 2-join-positions, there are $5^3 \times 4^2$ choices of codes, so there are $5^3 \times 4^2 \times 3= 6000$ 2-join codes.

So, all in all there are $3125 + 10000 + 6000 = 19125$ codes in the deluxe lock.

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    $\begingroup$ Your answer is correct $\endgroup$ Dec 24, 2014 at 9:02

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If $1-2$ is the same as $2-1$, there are only $5^3*10*4=5000$ 1-join codes.

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  • $\begingroup$ Thanks. Is the total therefore 3125 + 5000 + 6000/4 = 9625? $\endgroup$
    – oks
    Feb 13, 2015 at 10:02
  • $\begingroup$ I think so, yes. $\endgroup$
    – Empy2
    Feb 13, 2015 at 10:10

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