Question

The lab has now decided to require six scans in the passkey 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?___

Answer 1

Let the original sequence be T I M T R L.
Considering the above sequence, we have 3 cases
Case 1: When two scans misplaced are T’s.
Only one way possible T I M T R L as shifting places of T’s doesn’t make any difference.
Case 2: When one scan is T and other is one among M, I, R, L
One T can interchange place with M, I, R, L $\Rightarrow$ 4possible ways Since, there are 2 T’s, number of ways possible = 4 + 4 = 8
Case 3: When both the scans misplaced are not repeated figure. This case is similar to selecting 2 out of 4 different objects.
Hence, number of ways = ${}^4{C}_2=6$
Total number of arrangements possible = 1 + 8 + 6 = 15