now i am really tired of these permutation and combination

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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 vipulgoyal » Wed Jul 31, 2013 3:42 am
Again good Q
Favorable = Total - notfavorable
Using sets theory
6! - [ T@1 + J@2 + D@4 - T@1&D@4 - D@4&J@2 - T@1&J@2 + Tom, Jerry, and Donald@ 1,2&4th ]
(T or D or J) = T + D + J - (T ^ D) - (D ^ J) - (T ^ J) + (T ^ D ^ J)
5! + 5! +5! - 4! - 4! - 4! + 3!
360 - 72 + 6 = 294

720 - 294 = 426

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by GMATGuruNY » Wed Jul 31, 2013 2:47 pm
sana.noor wrote: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
This is a combinatorics problem that involves OVERLAPPING GROUPS.

Let T = arrangements with Tom in the wrong seat.
Let J = arrangements with Jerry in the wrong seat.
Let D = arrangements with Donald in the wrong seat.

Total BAD arrangements = T + J + D - (TJ + TD + JD) - 2(TJD).

The big idea with overlapping groups is to SUBTRACT THE OVERLAPS.
When we add together T + J + D (the arrangements with AT LEAST 1 person is in the wrong seat):
Arrangements with EXACTLY 2 people in the wrong seat -- TJ + TD + JD -- are counted twice, so these arrangements need to be subtracted ONCE.
Arrangements with ALL 3 people in the wrong seat -- TJD -- are counted 3 times, so these arrangements need to be subtracted TWICE.
By subtracting the overlaps, we ensure that no bad arrangement is overcounted.

T + J + D:
Here, AT LEAST 1 person must be in the wrong seat.
Number of options for the person in the wrong seat = 3. (Tom, Jerry, or Donald.)
Number of ways to arrange the remaining 5 people = 5! = 120.
To combine these options, we multiply:
3*120 = 360.

TJ + TD + JD:
Here, EXACTLY 2 people must be in the wrong seats.
From the trio of Tom, Jerry and Donald, the number of pairs that could be chosen to be put in the wrong seats = 3C2 = (3*2)/(2*1) = 3.
For the member of this trio who must sit in an ALLOWED seat, the number of options = 3. (Of the 4 remaining seats, any seat but the wrong seat.)
Number of ways to arrange the remaining 3 people = 3! = 6.
To combine these options, we multiply:
3*3*6 = 54.

TJD:
Here, Tom, Jerry and Donald must ALL be in the wrong seats.
Number of ways to arrange the remaining 3 people = 3! = 6.

Thus:
Total bad arrangements = 360 - 54 - 2(6) = 294.

Since the total number of ways to arrange the 6 people = 6! = 720, we get:
Good arrangements = total possible arrangements - bad arrangements = 720-294 = 426.

The correct answer is B.
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by Matt@VeritasPrep » Wed Jul 31, 2013 3:32 pm
How I'd do it:

Valid = Total - (Tom Wrong) - (Jerry Wrong) - (Donald Wrong) + (Tom and Jerry Wrong :D) + (Tom + Donald Wrong) + (Jerry and Donald Wrong) - (All 3 Wrong)

One nice thing about computing this way is that it's easy to remember what to do: start with the total, then SUBTRACT all the single violations, ADD BACK all the double violations, SUBTRACT all the triple violations, etc., alternating signs all the while, without having to remember much else.

Here we get 6! - 5! - 5! - 5! + 4! + 4! + 4! - 3! = 720 - 3*120 + 3*24 - 6 = 426.

Sana, you are going to be bummed out if there aren't any combo questions on your GMAT. These questions are not worth this level of attention.

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by vipulgoyal » Wed Jul 31, 2013 8:57 pm
T + J + D:
Here, AT LEAST 1 person must be in the wrong seat.
Number of options for the person in the wrong seat = 3. (Tom, Jerry, or Donald.)
Number of ways to arrange the remaining 5 people = 5! = 120.
To combine these options, we multiply:
3*120 = 360.

Hi Mitch, I think here should be Exactly rather then AT least