O/N 2010 P63 Q. no.: 6 (ii) and (iii)

[zeebujha]

http://www.xtremepapers.me/CIE/Internat ... _qp_63.pdf

Can anyone please explain me how I can approach this permutation question?? The mark scheme doesn't make sense!

 

[CaptainDanger]

ANyone?

 

[ahmed t]

well for the second part:
the three business men must sit in the first three seats which makes it 3! SINCE they can be moved about.
and the two married couples u have three couple of chairs and u need two so that would be 3P2 and within the couples the man and the woman can both be switched around so u times by 2 and u may change the smiths with the other couple eg, u may have the smiths in the second row and the others in the third row or vice versa, so u times by 2
and there are 5 windows left so it is 5! for the students
so it will be 3!* (3P2*2*2) *5!

as for the third part i have no clue

 

[zeebujha]

Thanks, I figured as much! What screwed me up totally was the third part!!!!

 

[ahmed t]

hey
if u get the anser for the third part kindly post it here thanks

 

[Amna]

for (iii):

If Mrs. Brown has to sit in the front row, she can sit in three possible places out of the 14 places available so we have 3/14

The five students can sit anywhere in the remaining 13 seats to we have 5/13

If Mrs. Lin has to sit behind a student... she can sit anywhere in the last three rows (=8 places) and two places in the second row... cuz in the third place Mrs. Lin would be in the front remember? so that's (8+2=10) places she can sit out of the 12 places left, or 10/12

So, we have 3/14 x 5/13 x 10/12 = 0.0687 which is the right answer...

 

[ManishAdhikari]

I think third part is a bit tough and I took 15 min to do myself before I looked at the answer. I did like if Mrs Brown sits in the front row then there are 10 possible positions where she can be behind a student including two of the second row and all the rest(not front) and one of the student must be in front of her. Now there are nine remaining people to be arranged which can be arranged in 11 places in 11!/2! ways. But there are 5 students and she can be behind any of them there must be 10*5*11!/2! ways that they can be arranged like this. Divide it by total possible arrangements for 12 and you have the answer.

 

[ManishAdhikari]

Sorry I made small mistake you will need to multiply first part by 3 because there are three seats in front row for Mrs Lin

 

[ahmed t]

is this an average question or is it hard?
if i couldnt do it first time is that a problem?

 

[IgoforA]

I managed to get a C for this paper...... Dude this paper's permutation and combination was freaking sizzling....

 

[zeebujha]

for (iii):

If Mrs. Brown has to sit in the front row, she can sit in three possible places out of the 14 places available so we have 3/14

The five students can sit anywhere in the remaining 13 seats to we have 5/13

If Mrs. Lin has to sit behind a student... she can sit anywhere in the last three rows (=8 places) and two places in the second row... cuz in the third place Mrs. Lin would be in the front remember? so that's (8+2=10) places she can sit out of the 12 places left, or 10/12

So, we have 3/14 x 5/13 x 10/12 = 0.0687 which is the right answer...

Why can Mrs. Lin sit anywhere in the last 3 rows? Suppose the first row has Mrs. Brown and 2 business men. The second row has Mr. Lin and Mr. Brown. That leaves only 1 place in the second row. So, if Mrs. Lin is to be behind a student, she cannot be on the second row but she can only be in one of the three places in the third row??