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24 May 2021, 11:59
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
63% (01:02) correct
37%
(01:05)
wrong
based on 182
sessions
History
Date
Time
Result
Not Attempted Yet
In how many different ways can you arrange 4 letters a, b, c, and d in a straight line so that 'b' comes before 'c'?
A. 2
B. 4
C. 6
D. 12
E. 14
A. 2
B. 4
C. 6
D. 12
E. 14
24 May 2021, 12:50
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paU1i
Note: b has to appear 'before' c, not necessarily 'immediately before'.
One line solution:
Total arrangements of a, b, c, d = 4! = 24
This allows for all letter to interchange positions among themselves in all possible ways. However, you DO NOT WANT b and c to interchange since b must appear before c. Thus, allowing the other letters to interchange in any manner possible, you simply want to CONSTRAIN the arrangements of b ad c which could have been done in 2! ways.
Thus, required number of ways = 4!/2! = 24/2 = 12
(Observe that it is similar to what would have happened if b and c were the same letter. Try to link these 2 concepts)
Answer D
Multi-line solution:
Possible cases:
b _ _ _ => c can be anywhere; hence interchange the 3 letters in 3! = 6 ways
_ b _ _ => 1st position is not c; hence, number of ways = 2 x 2 x 1 = 4 ways
_ _ b _ => 1st and 2nd positions are not c; hence, number of ways = 2 x 1 x 1 = 2 ways
Note: b cannot be in the last position since it has to be before c
Total = 6 + 4 + 2 = 12
Answer D
03 Jun 2023, 00:48
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The total no of letter is 4 and the required is 2 so
4C2 and then again we need the arrangement of those 2 letter
so we can write 4c2 as 4!/(4-2)!*2!=4*3*2!/2!*2!=12/2!=6 and there rearrangement of 2 letter so 6*2=12
2 method
As you know there is an arrangement of 2 letters that matter so directly use
4p2 = 4!/(4-2)!=4*3*2!/2!=12
Here you go
4C2 and then again we need the arrangement of those 2 letter
so we can write 4c2 as 4!/(4-2)!*2!=4*3*2!/2!*2!=12/2!=6 and there rearrangement of 2 letter so 6*2=12
2 method
As you know there is an arrangement of 2 letters that matter so directly use
4p2 = 4!/(4-2)!=4*3*2!/2!=12
Here you go
