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Tom, Jerry, and Donald and other three people sit in a line.

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

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New post 30 Jul 2008, 01:34
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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

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New post 30 Jul 2008, 02:15
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according to me answer is C - 432

here is my explanantion...

lets num the seats from 1 - 6

so we have seats 1 2 3 4 5 6

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

Nice question

Thanks

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Re: tom and jerry [#permalink]

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New post 30 Jul 2008, 02:26
GOOD LOGIC.. I DONT HAVE THE OA. BUT I WAS DOING THE SAME WAY, EXCEPT THAT I USED TO REPEAT 5! EVERY TIME . I HAD FORGOTTEN. PROB UR ANS IS OK.

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New post 30 Jul 2008, 02:49
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lets see ...

total number of ways = 720

number of ways when T:1 = 5! = 120
number of ways when J:2 = 5! = 120
number of ways when D:4 = 5! = 120

number of ways when T:1 and J:2 = 4! = 24
number of ways when J:2 and D:4 = 4! = 24
number of ways when D:4 and T:1 = 4! = 24

number of ways when t:1 and J:2 and D:4 = 3! = 6

we have to find when T:1 OR J:2 OR D:4 ....remember set theory
TUJUD = T + J + D - TiJ - JiD - DiT + TiJiD
= 120+120+120 - 24-24-24 + 6
= 294

so total number of ways when even one of these are not at the said positions = 720-294 = 426 ... option B

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New post 30 Jul 2008, 03:03
Hi durgesh79

Thanks for ur quicker and easier approach... i m confused can u point out where I am wrong...

thanks a lot

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New post 30 Jul 2008, 03:10
Good Explanation Durgesh +1

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Re: tom and jerry [#permalink]

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New post 30 Jul 2008, 05:12
durgesh79 wrote:
lets see ...

total number of ways = 720

number of ways when T:1 = 5! = 120
number of ways when J:2 = 5! = 120
number of ways when D:4 = 5! = 120

number of ways when T:1 and J:2 = 4! = 24
number of ways when J:2 and D:4 = 4! = 24
number of ways when D:4 and T:1 = 4! = 24

number of ways when t:1 and J:2 and D:4 = 3! = 6

we have to find when T:1 OR J:2 OR D:4 ....remember set theory
TUJUD = T + J + D - TiJ - JiD - DiT + TiJiD
= 120+120+120 - 24-24-24 + 6
= 294

so total number of ways when even one of these are not at the said positions = 720-294 = 426 ... option B



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

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New post 30 Jul 2008, 07:13
arjtryarjtry wrote:

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....

P(AuBuC)=P(A)+P(B)+P(C) - P(AnB) - P(AnC) - P(BnC) + P(AnBnC)

Last edited by durgesh79 on 30 Jul 2008, 10:38, edited 1 time in total.

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New post 30 Jul 2008, 07:17
sjgmat wrote:
Hi durgesh79

Thanks for ur quicker and easier approach... i m confused can u point out where I am wrong...

thanks a lot


the difference in your answer (432) and my answer (426) is 6... i guess you missed (or counted twice) TiJiD = 6

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New post 30 Jul 2008, 10:55
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..

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New post 30 Jul 2008, 11:09
arjtryarjtry wrote:
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

your equestion will become P+Q+R -A-B-C +x

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Re: tom and jerry   [#permalink] 30 Jul 2008, 11:09
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Tom, Jerry, and Donald and other three people sit in a line.

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