Question: The lab has now 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 if scans are allowed for any given person’s original scan?
Let us suppose original sequence is TIMTRL as indicated in the question. To answer this question, let us assume that all 6 scans in the sequence are distinct. As only 2 out of 6 scans can be out of place, it would imply that we choose 2 out of 6 scans and interchange their position. These can be done in 6 C2 or 15 ways. But since T occurs twice in the sequence, we eliminate the combination of T and T as interchanging these would not make any difference. So total number of ways possible where exactly 2 out of 6 scans can be in place and person can still be given access = 15 – 1 = 14
Another way the person can access is if the sequence of scans is as per the person’s original scan, which is possible in 1 way.