Mathematics: Post your doubts here!

[ZaqZainab]

Its statistics 1 by steve dobbs mixed exercise of permutations q19

can't find it
it says m!*n!=r!? that one

 

[Talha Irfan]

can't find it
it says m!*n!=r!? that one

definitely you're awesome

 

[Mayedah]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_32.pdf
Question 10 last part anyone ?

 

[ZaqZainab]

definitely you're awesome

I came across this
First, observe that 1!*n! = n!. This already solves the question (as the question does not say that m and n cannot be r).
e.g. 1! 20! = 20!
However, I guess this is not what the question is expecting.
Let's try to find m and n with the condition that none of them are 1.
Let x be a number.
Let y = x!.
Then y! = y . (y-1)!
= x! (y-1)!
This gives you an identity for finding products of factorials equal to other factorials.
(n!)! = n! (n! - 1)!
put n = 3 to get
6! = 3! 5!,
put n = 4 to get
24! = 4! 23!
you can put any n in this to get as many factorial products as you want.
However, still this only finds a specific type of factorial products. e.g. you cannot get 10! = 7! 6! from this.
So, let us put the condition that m and n should not be equal to r or (r-1) (as is happening above).
In this case, I have no idea on how to approach this.
http://mathworld.wolfram.com/FactorialPr...
Look at the last part of this page. This says that with these conditions and r > 10, r is atleast 18160 (and it doesn't give any such value of r either).
This suggests that this is "hard" to do (it may be easy but I don't know).
I believe your question is only expecting the (n!)! = n! (n! - 1)! idea, unless it is harder than I'm expecting it to be.

 

[daredevil]

salvatore
i tried that question using the method where u eliminate one variable each time and such...

but i'm not getting the right answer...

A star or Rutzaba can u guys help??

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_qp_33.pdf

Q6

 

[salvatore]

salvatore
i tried that question using the method where u eliminate one variable each time and such...

but i'm not getting the right answer...

A star or Rutzaba can u guys help??

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_qp_33.pdf

Q6

Lets hope we get some help before the exam

 

[abruzzi]

Here is my flimsy approach.
http://i.imgur.com/sAzYDKn.jpg

Thanks!

 

[Talha Irfan]

I came across this
First, observe that 1!*n! = n!. This already solves the question (as the question does not say that m and n cannot be r).
e.g. 1! 20! = 20!
However, I guess this is not what the question is expecting.
Let's try to find m and n with the condition that none of them are 1.
Let x be a number.
Let y = x!.
Then y! = y . (y-1)!
= x! (y-1)!
This gives you an identity for finding products of factorials equal to other factorials.
(n!)! = n! (n! - 1)!
put n = 3 to get
6! = 3! 5!,
put n = 4 to get
24! = 4! 23!
you can put any n in this to get as many factorial products as you want.
However, still this only finds a specific type of factorial products. e.g. you cannot get 10! = 7! 6! from this.
So, let us put the condition that m and n should not be equal to r or (r-1) (as is happening above).
In this case, I have no idea on how to approach this.
http://mathworld.wolfram.com/FactorialPr...
Look at the last part of this page. This says that with these conditions and r > 10, r is atleast 18160 (and it doesn't give any such value of r either).
This suggests that this is "hard" to do (it may be easy but I don't know).
I believe your question is only expecting the (n!)! = n! (n! - 1)! idea, unless it is harder than I'm expecting it to be.

Well, that is an intelligent approach. Weblink ?
But,
I didn't get the identity derivation
the identity in its final form was y! = x! (y-1)! where x and y are both different variables but the identity got a signle variable n . How is that possible?

 

[Crimson-Saint]

guys...i am doing AS maths and AL maths paper 12,62 and 32,72 in this may/june 14 session.......so i want to know how will i get me result?...like in the certificate will it be shown in my AS certificate? or will there be 2 seperate certificates like AL and AS?

 

[Beyond650]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_qp_33.pdf
Please please plaseee help me with qn 6(ii)
I'm freaking out

http://www.netcomuk.co.uk/~jenolive/vect18d.html

Summary:
1) Find the vector product of the two normals of the planes. The resultant vector will be the direction vector of the line of intersection of the two planes.
2) To find a point that lies on both planes let x = 0 and substitute into the equations for both planes. Solve the resultant equations simultaneously to get values of y and z. Then you have the values of the position vector x=0, y = -17 and z = -4.
3) By the way the direction vector is (1,7,2).

If you have trouble understanding then tell me and I will type out the solution in word.

 

[Rutzaba]

Lets hope we get some help before the exam

wait n pray

 

[Rutzaba]

Lets hope we get some help before the exam

see now u may not always get the ryt answer... but it must have the same ratio as the b1 part of the line equation
maybe they will give the answer 1;2;3
where as u will be getting the answer
2'4'6
although these r different they must be in the same ratio
the answer is still accepted

 

[Rutzaba]

guys...i am doing AS maths and AL maths paper 12,62 and 32,72 in this may/june 14 session.......so i want to know how will i get me result?...like in the certificate will it be shown in my AS certificate? or will there be 2 seperate certificates like AL and AS?

private or school?

 

[salvatore]

http://www.netcomuk.co.uk/~jenolive/vect18d.html

Summary:
1) Find the vector product of the two normals of the planes. The resultant vector will be the direction vector of the line of intersection of the two planes.
2) To find a point that lies on both planes let x = 0 and substitute into the equations for both planes. Solve the resultant equations simultaneously to get values of y and z. Then you have the values of the position vector x=0, y = -17 and z = -4.
3) By the way the direction vector is (1,7,2).

If you have trouble understanding then tell me and I will type out the solution in word.

thanks a lot.. the site gives a crystal clear explanation
daredevil

 

[salvatore]

wait n pray

see now u may not always get the ryt answer... but it must have the same ratio as the b1 part of the line equation
maybe they will give the answer 1;2;3
where as u will be getting the answer
2'4'6
although these r different they must be in the same ratio
the answer is still accepted

Thanks, I found the solution.
Please remember me in your prayers

 

[hope4thebest]

Can they ask for perpendicular distance between two planes? If yes what's the procedure than?

 

[Rutzaba]

t

Lets hope we get some help before the exam

this is easy se the equationsthe y is already eliminating
adding the rest u get
4x-2z=8
x=8+2z/4
next to eliminate z multiply eq 1 with 2
6x-2y+4z=18
x+y-4z=-1
7x-y=17
x=17+y /7
x=x=x
(x-0)/1 onlyway it can remain x
(x-0)/1 =(y+17)/7 = 2(z+4)/4
(x-0)/1 =(y+17)/7 = (z+4)/2

 

[Rutzaba]

srry had to go. i sure will buddy u too

 

[Beyond650]

Can they ask for perpendicular distance between two planes? If yes what's the procedure than?

Not in the syllabus. The only distance-pertaining questions will be to:
"find the perpendicular distance from a point to a plane, and from a
point to a line"
Resources for the above syllabus point:
http://members.tripod.com/vector_applications/tutorials/index.html
http://www.vitutor.com/geometry/distance/distance_lines.html

 

[Rutzaba]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_32.pdf
Question 10 last part anyone ?

Can they ask for perpendicular distance between two planes? If yes what's the procedure than?

http://wwwf.imperial.ac.uk/metric/m...rdinate_geometry/distance_between_planes.html
not the same question but it will help
http://www.math.ucla.edu/~ronmiech/Calculus_Problems/32A/chap11/section5/718d65/718_65.html