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very well explained.... thankyou sooooo much littlecloud11Okay, this is a tricky one.
View attachment 25529
See the picture. The red seats highlight the ones that Mrs. Lin can occupy. She has to sit directly behind a student, so if she occupies those seats then it's possible for a student to be placed in front of her. The reason I kept one seat in the second row uncolored is because one of the seats in the front row is occupied my Mrs. Brown. A student can only take two seats in the front row.
So for Mrs. Lin the possible combinations are 10P1
Mrs. Brown has to sit in one of the front seats so 3P1
She can sit behind any of the 5 students, so it's 5P1, the reason you don't count the possible seats for the student is because it doesn't matter where they are placed as long as one of them is in front of Mrs. Lin
And finally after Mrs. Lin, a student and Mrs. brown has been placed there are 9 remaining passengers and 11 remaining seats. There is no restrictions. So it's 11P9
Total no. of ways = 10P1 * 3P1* 5P1 * 11P9
If all 12 passangers are seated randomly = 14P12
so probability = 10P1 * 3P1* 5P1 * 11P9/ 14P12 =.0687
Hope you get this!
Okay, this is a tricky one.
View attachment 25529
See the picture. The red seats highlight the ones that Mrs. Lin can occupy. She has to sit directly behind a student, so if she occupies those seats then it's possible for a student to be placed in front of her. The reason I kept one seat in the second row uncolored is because one of the seats in the front row is occupied my Mrs. Brown. A student can only take two seats in the front row.
So for Mrs. Lin the possible combinations are 10P1
She can sit behind any of the 5 students, so it's 5P1, the reason you don't count the possible seats for the student is because it doesn't matter where they are placed as long as one of them is in front of Mrs. Lin
Mrs. Brown has to sit in one of the front seats so 3P1
And finally after Mrs. Lin, a student and Mrs. brown has been placed there are 9 remaining passengers and 11 remaining seats. There is no restrictions. So it's 11P9
Total no. of ways = 10P1 * 3P1* 5P1 * 11P9
If all 12 passangers are seated randomly = 14P12
so probability = 10P1 * 3P1* 5P1 * 11P9/ 14P12 =.0687
Hope you get this!
ok guys heres the deal
6 ii 5 students sit anywhere by the window... 5! =120
3 business men sit anywhere= 3!= 6
now for the married couples... there are 2 couples =2 two seats each =2 and 3 seats remaining for each couple =3p2
3p2 *2*2= 24
5!=120
3!=6
24*6*120 = 17280
Some help with Part ii as well.. Since I don't understand why the mark scheme is ignoring the other window seats... and just doing a 3P3*4P4*5P5 ..
Would really appreciate it ...
Thanks
Thanks for solving all our problems....View attachment 25598
The students cannot take the window seats in the first row because the 3 business people have to be seated there. Each married couple must be seated in a different row, but they have to be on the same side of the isle. So one of the couples have to take two window seats so that both couples are in different rows. So you see, only 5 window seats remain, and the students have to take these 5 seats in any order. So it's 5! and not 9P5.
Thanks for solving all our problems....
bother? you aren't bothering meSorry to bother you but can you please solve it for me ? I didn't get you..
View attachment 25598
The students cannot take the window seats in the first row because the 3 business people have to be seated there. Each married couple must be seated in a different row, but they have to be on the same side of the isle. So one of the couples have to take two window seats so that both couples are in different rows. So you see, only 5 window seats remain, and the students have to take these 5 seats in any order. So it's 5! and not 9P5.
well this was a fun questionhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_qp_4.pdf
@Obsidian Fl1ght
@Pie-man
its question 6(iii) ?
JazakAllah
isn't this a MATHEMATICS thread?can someone please explain to me the link between work done by driving force, work one by resistance, k.e, p.e
isn't this a MATHEMATICS thread?
the link from the rule "Final Energy = Initial Energy + Driving force - Resistance force" ? or the one "Driving force - Resisting force = mass x acceleration"?can someone please explain to me the link between work done by driving force, work one by resistance, k.e, p.e
good luck with that O-Level seems like a piece of cake after taking the ASOmg! I just scrolled up and found out this is for A levels! MY BAD! HAHA!
LMAO! Next year I'll be posting on this page!
well i'm doing the AS-Level but i'll ask one of my friends ... when any of them wake up [i woke up early to study]anyone please??
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