Six alphabets A B C D, have to be arranged in six numbered positions (1-6). How many ways can you arrange them so that neither A is in the position numbered 1, nor B is in position numbered 2 and C is in position numbered 4.
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Good Question on P and C
Source: Beat The GMAT — Problem Solving |
Shouldn't 426 be the answer!!
1. A should not be in position number 1.
No. of ways in which A occurs in position 1 = 5! = 120 arrangements.
Therefore possible arrangements = 6! - 120 = 600 arrangements.
2. B should not be in position number 2.
No. of ways in which B occurs in position 2 = 5! = 120 arrangements.
But in this case, A occurs in position 1 by 24 ways which is already included in the first case.
Therefore possible arrangements after including second constraint = 600-120 + 24 = 504 arrangements.
3. C should not be in position 3:
No. of ways in which C occurs in position 3 = 5! = 120 arrangements.
In this case also A occurs in position 1 by 24 ways which is already included in the first case.
Also B occurs in position 2 by 18 ways (actually it's also 24 ways but ABC can occur in 6 ways which is already included), which is already included in the second case.
Therefore total arrangements after considering all constraints = 504-120+24+18 = 426
I hope that the answer is correct. comments please
1. A should not be in position number 1.
No. of ways in which A occurs in position 1 = 5! = 120 arrangements.
Therefore possible arrangements = 6! - 120 = 600 arrangements.
2. B should not be in position number 2.
No. of ways in which B occurs in position 2 = 5! = 120 arrangements.
But in this case, A occurs in position 1 by 24 ways which is already included in the first case.
Therefore possible arrangements after including second constraint = 600-120 + 24 = 504 arrangements.
3. C should not be in position 3:
No. of ways in which C occurs in position 3 = 5! = 120 arrangements.
In this case also A occurs in position 1 by 24 ways which is already included in the first case.
Also B occurs in position 2 by 18 ways (actually it's also 24 ways but ABC can occur in 6 ways which is already included), which is already included in the second case.
Therefore total arrangements after considering all constraints = 504-120+24+18 = 426
I hope that the answer is correct. comments please
