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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) where at most two scans (Out of six) are Out of place, 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 for any given person’s original scan?

 

Option: 1

15


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

The question asks us to find how many different sequences of scans are allowed for any given person's original scan, given that the lab has decided to allow variations of the original sequence where at most two scans (out of six) are out of place, as long as the finger originally scanned twice is scanned twice and the other fingers are scanned once.

We can use the following cases to solve the problem:

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

Case 2: Exactly two scans are 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 two scans that are out of place in (26?) =15 ways, but we need to divide by 2 to account for the fact that the order in which the scans are out of place doesn't matter. This gives us 2(26?) × 2!?=14 sequences.

Therefore, the total number of allowed sequences is 1+14=15?

Here is a table that summarizes the two cases:

Case

Number of sequences

None of them misplaced

1

Exactly two are misplaced

14

Total

15

 

Posted by

Rakesh

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