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  • A high security research lab requires the researchers to set a pass key sequence based on the scan of the five fingers of their left hands. When an employee first joins the lab, her fingers are sca

A high security research lab requires the researchers to set a pass key sequence based on the scan of the five fingers of their left hands. When an employee first joins the lab, her fingers are scanned in an order of her choice, and then when she wants to re-enter the facility, she has to scan the five fingers in the same sequence.
The lab authorities are considering some relaxations of the scan order requirements, since it is observed that some employees often get locked-out because they forget the sequence

Question : The lab has flow decided to require six scans in the pass key sequence, where exactly one finger is scanned twice, and the other fingers are scanned exactly once, which can be done in any order For example, a possible sequence is TIMTRL.
Suppose the lab allows a variation of the original sequence (of six inputs) so that input in the form of scanned sequence of six fingers is allowed to vary from the original sequence by one place for any of the fingers, as long as the finger originally scanned twice is scanned twice and other fingers are scanned once.
How many different sequences of scans are allowed if the original scan sequence is LRLTIM?

Option: 1

8


Option: 2

11


Option: 3

13


Option: 4

14


Answers (1)

The question asks us to find how many different sequences of scans are allowed if the original scan sequence is LRLTIM, given that the lab has decided to allow variations of the original sequence where at most one scan is out of place.

We can use the following cases to solve the problem:

Case 1: None of the scans are out of place.

Case 2: Exactly one scan is out of place.

For Case 1: There is only one way to do this since the finger originally scanned twice must be scanned twice and the other fingers must be scanned once.

For Case 2: We can choose any one scan that is out of place in 5 ways, but we need to multiply by 2 to account for the fact that the order in which the scans are out of place doesn't matter.

Therefore, the total number of allowed sequences is 1+5×2=13?

Here is a table that summarizes the two cases:

Case

Number of sequences

None of them misplaced

1

Exactly one is misplaced

10

Total

13

The 5 scans that can be out of place are LR, RL, LT, TI, and IM. For each of these scans, there are 2 ways to move it to a different position since the order in which the scans are out of place matters.

For example, if the original sequence is LRLTIM, and we move the scan LR to the first position, we get the sequence RLTLMI. 

We can also move the scan LR to the third position, to get the sequence LTIMRL.

The other 4 scans can be moved to different positions in a similar way.

Therefore, there are a total of 5 x 2 = 10 sequences that can be formed by moving one scan out of place.

In addition to the 10 sequences that we have just found, there is also the original sequence, LRLTIM.

Therefore, the total number of allowed sequences is 10 + 1 = 13?

 

Posted by

Ramraj Saini

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