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Tom, Jerry, and Donald and other three people sit in a line. [#permalink]

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30 Jul 2008, 01:34

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

now without any constraint total possible combinations are

6! -- (a)

now lets start with 1st contraint -

lets say Tom (T) is on 1

then remaining seats can be occupied in 5! ways (b)

now lets take 2nd constraint

lets say Jerry (J) is on 2

( Note possible combination is less than 5! as while assuming that T is on 1 we have already taken the case where J was on 2 so we have to consider only remaining cases i.e when T is not on 1 )

so total combinations possible are - 4 ( excluding T on 1) x 1 x 4 x 3 x 2 x 1= 4*4! ... (c)

now lets take 3rd constraint

lets say Donald (D) is on 4

( Note possible combination is less than (c) as while assuming that T is on 1 and J on 2 we have already taken the case where D was on 4 so we have to consider only remaining cases i.e when T is not on 1 and J is not on 2 )

so total combinations possible are - 4 (excluding T on 1) x 3 ( excluding J on 2 ) x 3 x1 x 2 x 1 = 12 * 3! .. (d)

final required soln is a - b-c-d

which we get as 432

please let me know if it is correct and also if there is better way of solving the prob

I THINK ITS -2(all 3)....i.e-2(TiJiDi)... is it not???

lets get back to basics .... (AnB is 'A intersection B' and AuB is 'A union B')

AuBuC should include everying that is in A, B and C and common areas should be counted only once...

AnB is a part of A AnB is a part of B

so when we write A+B+C we are actully counting AnB twice .... so we have to substratc AnB once to get the final answer ... similar logic for AnC and BnC ....

AnBnC is a part of A, B, C, AnB, AnC and BnC.... so when we add A+B+C and substract - AnB-AnC-BnC we are adding AnBnC 3times and substraticg it 3 times .... so we have to add it only once to get the final answer....

HMMM, MY thinking is like this... if P is all people of grp 1(say) if Q is all people of gp 2 if R is all people of gp 3, let common bet'n 1 and 2 be a, betn 2 and 3 be b and betn 3 and 1 be c. let area of all the three be x. then people of [ONLY 1= P-x-a-c ;people of ONLY 2 =Q-x-a-b, .... and ONLY 3=R-x-b-c..]... these are single users two users= a+b+c three users = x. on adding the three, P-x-a-c+Q-x-a-b+R-x-b-c+a+b+c+2x = P+Q+R-(a+b+c)-2x... is there something wrong here... im unable to get it..

HMMM, MY thinking is like this... if P is all people of grp 1(say) if Q is all people of gp 2 if R is all people of gp 3, let common bet'n 1 and 2 be a, betn 2 and 3 be b and betn 3 and 1 be c. let area of all the three be x. then people of [ONLY 1= P-x-a-c ;people of ONLY 2 =Q-x-a-b, .... and ONLY 3=R-x-b-c..]... these are single users two users= a+b+c three users = x. on adding the three, P-x-a-c+Q-x-a-b+R-x-b-c+a+b+c+2x = P+Q+R-(a+b+c)-2x... is there something wrong here... im unable to get it..

in you definition on a, b and c you are counting common elements only between two sets .... 1 intersection 2 means all elements common between 1 and 2, this will also include elements which are common between 1, 2 and 3.

so if A = 1 intersection 2 A = a + x => a = A-x B = b + x => b = B-x C = C + x => c = C-x