Bunuel
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) 426
(B) 432
(C) 438
(D) 444
(E) 720
Directly counting the arrangements will be cumbersome. Think why.
Say there are 5 options for first seat (excluding Tom).
Now the number of options for second seat will depend on who takes the first seat. If Jerry does then there are 5 options, but if someone else does, then there are 4 options.
and so on...
So instead, focus on trying to get the unacceptable arrangements.
Total arrangements of 6 people on 6 seats = 6! = 720
Unacceptable arrangements:
Tom sits on the first seat = 5!
Jerry sits on the second seat = 5!
Donald sits on the fourth seat = 5!
But note that there is double counting here - When Tom sits on the first seat, there are 5! ways of arranging others including ways in which Jerry sits on the second seat and/or Donald sits on the fourth seat. These ways are counted again in the next 5!
It becomes a sets question. There is overlap in these sets of 5! each. Think of your venn diagram of three overlapping sets. There are 4! cases in which Tom and Jerry both are sitting on unacceptable seats. Similarly for Donald too. There are 3! cases in which all three are sitting on unacceptable seats.
So Total unacceptable cases = 5! + 5! + 5! - 4! - 4! - 4! + 3! = 3*5! - 3*4! + 3! = 3! * (60 - 12 + 1) = 294
(Using the formula n(Total) = n(A) + n(B) + n(C) - n(A and B) - n(B and C) - n(C and A) + n(A and B and C)
Acceptable cases = 720 - 294 = 426
Answer (A)