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by vaivish » Fri Aug 01, 2008 2:07 am
Tom, Jerry, and Donald and other three people sit in a line. From left to right, if Tom cannot sit on the first seat, Jerry cannot sit on the second seat, and Donald cannot sit on the fourth seat, hoe many different arrangements are possible?
(A) 720
(B) 426
(C) 432
(D) 438
(E) 444


OA is B
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by parallel_chase » Fri Aug 01, 2008 5:41 am
The answer I am getting is 400

5*5*4*2*2

we have total of 6 people. Let me know where i went wrong.
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by sudhir3127 » Fri Aug 01, 2008 6:13 am
My answer is 426..

here it goes..

total number of arrangements using 6 persons is 6!= 720 ways,.
Tom cant be in first postion

therefore possible arrangements are 6!-5! = 600 ways

Jerry cant be position 2 in 5! = 120 ways

Tom occurs in position 1 in 24 (4!) ways which already included in the first case above

therefore the arrangements are 600-120+24 = 504 ways ( 600- (5!-4!) )

Donald cant be 3rd postion in 5! = 120 ways
Bu A occurs in postion 1 in 24 (4!) ways and B in position 2 in 18 ways ( 4!-3!)

there fore possible arrangement are 504-120+24+18 =426 ways..


hope its clear,,,
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by sudhir3127 » Fri Aug 01, 2008 8:29 am
Any takers for this .. though i got it right ...i found it to be a very long and tedious procedure wherein anyone can go wrong during the exam tension..

Any other shorter method on this will be highly Appreciated!!!
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