Number of possible movements

Question

In a certain exercise of a previous exam, there was a question introduced regarding a safe-deposit box. The "password" or combination of movements that open the box consist on the following: there are $10$ different positions disposed on a circle, and a combination is a list of $5$ movements which consist on spinning the circle so that after those $5$ movements, the final position is the same as the initial one. It has to be taken into account that no movement can consist on doing a full rotation of the circle or wheel, neither consist on not doing anything. I am asked to find how many of these "passwords" are there. Here is my attempt:
Since the positions are distinguishable, one first chooses a position: $10$ options. After that, one can consider the variables $x_1,x_2,x_3,x_4,x_5$, where $x_i$ represents the amount of positions spinned in the $i$-movement. Therefore, since the initial position has to be the same as the final one, the sum of positions spinned has to be $10$, and note that $x_i \geq 1$ since no movement can be "rotating 0 positions". Therefore, and since each movement is distinguishable (the order matters as different order means different "password"), it is the same as counting the number of solutions of $x_1 + x_2 + x_3 + x_4 + x_5 = 10, x_i \geq 1$, which is $9 \choose 4$. Therefore, using the product rule, there are $10 \cdot {9 \choose 4}$ possible "passwords". Is this right?

Answer 1

there are 10 different positions disposed on a circle, and a combination is a list of 5 movements which consist on spinning the circle so that after those 5 movements, the final position is the same as the initial one. It has to be taken into account that no movement can consist on doing a full rotation of the circle or wheel, neither consist on not doing anything.

First of all, I disagree with the posting's analysis that

since the initial position has to be the same as the final one, the sum of positions spinned has to be $10$.

For example, consider the (valid) combination represented by 1-9-2-8-3-1, where the initial position and final position are both $~1.~$

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Index the positions by the elements of set $~A = \{1,2,\cdots,10\}.$ You can assume that if $~a,b \in A ~: ~a \neq b,$ and if the $~i$-th position is $~a,~$ and the $~(i+1)$-th position is $~b~$ that (in effect) the corresponding movement was from $~a~$ directly to $~b.~$

Note that in the real world, the previous paragraph would be construed as ambiguous. For example, if the clockwise direction represents that the positions are in increasing order, and if the initial position is $~1,~$ and the position after the first movement is $~2,~$ then going from $~1~$ directly to $~2~$ could be construed as either traveling clockwise along the circle, or counter-clockwise along the circle.

For simplicity, I am going to assume that a combination is completely defined by the the set of successive positions represented by the combination, and that when going from one position to another, it is deemed irrelevant whether you travel in a clockwise or counter-clockwise direction.

Note that if the assumption in the previous paragraph is wrong, the remedy is to simply apply the scalar of $~2^5,~$ since there will always be two possible directions (i.e. clockwise or counter-clockwise) that one can take to make a movement.

So, I am assuming that the problem is asking for an enumeration of how many ordered $~6$-tuples that there are where:

• Each of the $~6~$ components of the ordered $~6$-tuple is an element in $~A.$

• For $~i \in \{1,2,3,4,5\},~$ component $~i~$ is not equal to component $~i+1.$

• Component $~6~$ equals component $~1.$

Normally, I would use Inclusion-Exclusion against a problem like this, as discussed here and here.

The following discussion illustrates a trap in the problem, which makes both Inclusion-Exclusion and a (vanilla) direct approach problematic.

Suppose that the first component of the ordered $~6$-tuple is $~1.$

Then, you can not have any satisfying ordered $~6$-tuple where the fifth component is also $~1.~$ This implies that if the $~4$-th component is $~1,~$ then you have $~9~$ choices for the $~5$-th component. Alternatively, if the $~4$-th component is not $~1,~$ then you have only $~8~$ choices for the $~5$-th component.

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To conquer this problem, I am going to re-define the problem, and then use a (somewhat convoluted) form of recursion. First of all, it will be assumed that the first component of the ordered $~k$-tuple is $~1.~$ Then, at the end of the problem, a scalar of $~10~$ will be applied, which represents that there are $~10~$ choices for the first component.

Instead of a combination being represented by a satisfying ordered $~6$-tuple, assume that a combination is represented by a satisfying ordered $~k$-tuple, where $~k \in \{3,4,5,6\},~$ and let $~f(k)~$ denote the number of satisfying ordered $~k$-tuples.

Then, the problem reduces to computing $~f(6).$

Note, that unless otherwise specified, it is assumed that the first component of the ordered $~k$-tuple is $~1.$

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To compute $~f(3),~$ note that you have

• $~9~$ choices for the second component,

• and then $~1~$ choice for the third component.

So,

$$f(3) = 9.$$

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To compute $~f(4),~$ note that you have

• $~9~$ choices for the second component.
  At this point, you are guaranteed that the second component is $~\neq 1.$

• So, then you have $~8~$ choices for the third component.
  That is, the third component must be different from the second component.
  
  Further, the third component must also be $~\neq 1,~$ since [when computing $~f(4)$] the fourth component must be both different from the third component and equal to $~1.$

• Then, you will have $~1~$ choice for the fourth component.

Therefore,

$$f(4) = 9 \times 8 \times 1 = 72.$$

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Now, $~f(6)~$ can be computed.

Either the third component will match the first component, or it won't.

• Case 1
  The third component matches the first component.
  There are $9~$ ways that this can occur. At this point, the third component is $~1,~$ and $~f(4)~$ has been computed to be $~72.$ This means that there are $~72~$ ways of completing the ordered $~6$-tuple so that the third and sixth components are both $~1.$
  
  Therefore, the partial sum for Case 1 is
  $9 \times 72 = 648.$

• Case 2
  There are $~9 \times 8 = 72~$ ways that the first and third component can be different. At this point you have to distinguish whether the 4th component is identical to the first component.

• Case 2.1
  There are $~72 \times 1 = 72~$ ways that the fourth component can match the first component, under the assumption that the third component was different from the first component.
  
  When the fourth component matches the first component, then the number of ways of completing the ordered $~6$-tuple corresponds to $~f(3) = 9.$
  
  Therefore, the partial sum for Case 2.1 is
  $72 \times 9 = 648.$

• Case 2.2
  There are $~72 \times 8 = 576~$ ways that the fourth component can be different from the first component, under the assumption that the third component was different from the first component.
  
  When the fourth component does not match the first component, then there are only $~8~$ ways of completing the ordered $~6$-tuple, since the fifth component must be different from both the fourth component and the first component.
  
  Therefore, the partial sum for Case 2.2 is
  $576 \times 8 = 4608.$

So,

$$f(6) = 648 + 648 + 4608 = 5904.$$

Consequently, since the scalar of $~10~$ must be applied to compensate for there being $~10~$ choices for the first component, the number of satisfying combinations (i.e. ordered $~6$-tuples) is

$$5904 \times 10 =59040.$$

Answer 2

Cogito

5-number numerical password. 1st number = Last number.

Total number of possible keys = $10^4 = 10000$

However for 1 of these keys, $2^5 = 32$ different movements.

Number of movements = $32 \times 10000 = 320000$