Question

How many different sequences of scans are allowed for any given person's original scan?

• A. 7
• B. 5
• C. 8
• D. 13
• A. 8
• B. 11
• C. 13
• D. 14

Answer 1

Correct Answer: 11

Since the order of exactly one out of the five scans can’t be changed, either all the scans are in the correct order or one pair of scans can be varied, i.e. their positions can be interchanged.
Case (1): when all the scans are in the correct order = 1 way
Case (2): when exactly two are interchanged:
We can choose any two of the five scans that can be interchanged in 5C2 ways, viz. 10
Both case (1) and case (2) together = 11.
Ans: (11)

Answer 2

The correct answer is (C):
Let the original scan be: TIMRL
(1) All sequence as original = 1 way
(2) Interchange of TI = 1 way
(TI) + (RL) = 1 way
→ 2 way
(3) Interchange of IM = 1 way
(IM) + (RL) = 1 way
→2 way
(4) Interchange of MR = 1 way
(MR) + (TI) = 1 way
→2 way
(5) Interchange of RL = 1 way
Total = 1 + 2 + 2 + 2 + 1 = 8 ways.

Answer 3

Let us say original input: TIMTRL.
Case (1): None of them misplaced : 1.
Case (2): When exactly two are misplaced.
T can be misplaced →4 ways.
I can be misplaced →4 ways.
M can be misplaced →3 ways.
T can be misplaced →2 ways.
R can be misplaced →1 way.
Total ways in case (2) = 4 + 4 + 3 + 2 + 1 = 14 ways.
Both case (1) and case (2) = 14 + 1 = 15 ways.
Ans:- 15

Answer 4

The correct answer is (C):
Given LRLTIM
The distinct possibilities are:
1. No shift = 1 way
2.(a) LR = 1 way
(b) LR + LT = 1 way
(c) LR + LT + IM = 1 way
(d) LR + IM = 1 way
(e) LR + IT = 1 way (Total 5 ways)
3.(a) RL = 1 way
(b) RL + TI = 1 way
(c) RL + IM = 1 way (Total 3 ways)
4.(a) LT = 1 way
(b) LT + IM = 1 way (Total 2 way)
5. TI = 1 way
6. IM = 1 way
Total ways = 1 + 5 + 3 +2 +1 + 1 = 13 ways.