Author Topic: Additional Math Help HERE ONLY...!  (Read 75472 times)

Offline shan2391

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Re: Additional Math Help HERE ONLY...!
« Reply #105 on: June 02, 2009, 07:48:11 pm »
I have prob solving velocity n relative velocity and is there any esy way to solve permutation Q.

Offline vanibharutham

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Re: Additional Math Help HERE ONLY...!
« Reply #106 on: June 02, 2009, 07:48:47 pm »
Here's a common Permuatations and Combinations Questions:

a) The producers of a play require a total cast of 5, of which 3 are actors and 2 are actresses. He auditions 5 actors and 4 actresses for the cast. Find the total number of ways in which the cast can be obtained.

b) Find how many different odd 4-digit numbers LESS than 4000 can be made from the digits 1,2,3,4,5,6,7 if no digit may be repeated.



The FIRST thing to consider when solving Permutations and Combinations questions is recognizign what the hell you are doing...

ALWAYS REMEMBER:
PERMUTATIONS HAVE TO DO WITH ARRANGEMENTS OF THINGS
COMBINATIONS HAVE TO DO WITH HOW MANY POSSIBILITIES ARE THERE, WITHOUT WORRYING ABOUT ARRANGEMENTS


if we follow that rule...
If for example, we had three things.... A,B,and C....

the combination ABC would be the same as BAC or CAB or BAC etc.
However the permutations are different...


Back to the question...
a) The producers of a play require a total cast of 5, of which 3 are actors and 2 are actresses. He auditions 5 actors and 4 actresses for the cast. Find the total number of ways in which the cast can be obtained.

This is obviously worried about combinations, not arrangements...

This my method of solving this question:

From:     5 actors       4 actresses
Choose:  3 actors       2 actresses

=> 5C3 x 4C2 = 10 x 6 = 60 combinations


b) Find how many different odd 4-digit numbers LESS than 4000 can be made from the digits 1,2,3,4,5,6,7 if no digit may be repeated.

Obviously this is a permutation question :) its got to with arrangements....

This is my method in solving this;

First, find out how many 4-digit odd numbers you can have...

well

_ x _ x _ x _

We have 4 digits to fill from 7 numbers.... however for it to be an odd number, the last digit NEEDS to be either 1,3,5 or 7...

and that gives us four digits on the last one....
_ x _ x _ x 4

What about the rest.... well.....
Obviously you will only have 6 choices for your first one, 5 for the next, and 4 for the next...
which gives us

6 x 5 x 4 x 4 = 480 odd numbers....


To find out how numbers from those odd numbers are LESS than 4000...
THEY NEED TO START WITH A 1, 2 or a 3... . and end with a 1,3,5 or 7...

First lets see all possibilites starting with 2....

Obviously because its starting with 2, we only have 1 option for the start.
1 x _ x _ x _

once again, we have 1,3,5,7 for the last option.... that gives us ... 4

1 x _ x _ x 4

how many numbers do we have left... 5 to choose from...

therefore.... 1 x 5 x 4 x 4 = 80 odd numbers Starting with 2, and that are less than 4000....


Now for numbers starting with 1 or 3....

we have two options for the first one....

2 x _ x _ x _

and we only have 3 options for the last one.... 1,3,5,7 but because we are already using one of them as our first number, we can only choose 3....

2 x _ x _ x 3

how many numbers do we have left? well we have.... 5...

therefore...

2 x 5 x 4 x 3 = 120....


120 + 80 = 200

We have 200, four digit odd numbers less than 4000...


Question 5 from October/November 2002 Paper 2...Cambridge

Read that for permutation
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Offline vanibharutham

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Re: Additional Math Help HERE ONLY...!
« Reply #107 on: June 02, 2009, 07:51:52 pm »
Relative velocity questions are bugging me as well

However, to solve them , you need to break them down logically...

Relative velocity literally means velocity relative to another....
If they give you an airplane and wind question, you will basically need to know that

V(plane) = V(plane in still air) + V(wind)

This statement holds true if you think it through logically...
In still air, there is no wind, thereforoe your plane will have a certain velocity
However when the wind vector has been added, you need add that to your velocity in still air to calculate your overall velocity of the plane
A genius is 1% intelligence, 99% effort.

Offline shan2391

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Re: Additional Math Help HERE ONLY...!
« Reply #108 on: June 02, 2009, 08:25:47 pm »
Thanks vanibharutham
By the way giv a shot at this 1
Q7(iii) Nov 08 p1

Offline IO4567

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Re: Additional Math Help HERE ONLY...!
« Reply #109 on: June 02, 2009, 08:26:47 pm »
hey i just have a question out of curiosity: what is additional maths? is it like harder maths or something?
thank you and good luck!

Offline vanibharutham

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Re: Additional Math Help HERE ONLY...!
« Reply #110 on: June 02, 2009, 08:30:37 pm »
it covers most of the A-level Core 1 and Core 2 maths i think

and our school let us take it because we finished our IGCSE and GCSE maths a year earlier than normal... its much harder
A genius is 1% intelligence, 99% effort.

Offline shan2391

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Re: Additional Math Help HERE ONLY...!
« Reply #111 on: June 02, 2009, 09:45:46 pm »
hey vanibharutham
Q.10 is quite tricky 4 me can u giv it a shot.
paper attached

Offline ineedaz

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Re: Additional Math Help HERE ONLY...!
« Reply #112 on: June 02, 2009, 09:46:29 pm »
Particle mechanics and vector questions are my worst enemies! oh and permutations...i thnk combinations are ok!! anyone know the grade thresholds, by any chance??

Offline shan2391

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Re: Additional Math Help HERE ONLY...!
« Reply #113 on: June 02, 2009, 09:50:13 pm »
Particle mechanics and vector questions are my worst enemies! oh and permutations...i thnk combinations are ok!! anyone know the grade thresholds, by any chance??
ya approx 80% or higher will earn u an A

Offline ineedaz

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Re: Additional Math Help HERE ONLY...!
« Reply #114 on: June 02, 2009, 09:54:59 pm »
Particle mechanics and vector questions are my worst enemies! oh and permutations...i thnk combinations are ok!! anyone know the grade thresholds, by any chance??
ya approx 80% or higher will earn u an A
cool...wat about an A*?? thats wat im aiming for...

Offline shan2391

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Re: Additional Math Help HERE ONLY...!
« Reply #115 on: June 02, 2009, 09:57:40 pm »
hey vanibharutham
Q.10 is quite tricky 4 me can u giv it a shot.
paper attached

any 1 please solve the Q.

Offline ineedaz

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Re: Additional Math Help HERE ONLY...!
« Reply #116 on: June 02, 2009, 10:08:46 pm »
hey vanibharutham
Q.10 is quite tricky 4 me can u giv it a shot.
paper attached

Q10
(i) Let the velocity be = xi+yj
so:
x^2 + y^2 = (15(2)^0.5)^2 = 450
but the angle between x and y is 45 because of the (2)^0.5
so x = y
therfore 2(x)^2 = 450
x^2 = 225
x = 15
therfore the velocity = 15i+15j
MAke sense so far??

Offline shan2391

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Re: Additional Math Help HERE ONLY...!
« Reply #117 on: June 02, 2009, 10:19:44 pm »
Got it man it was north east thats why it is 45deg
Thanks
« Last Edit: June 02, 2009, 10:28:58 pm by shan2391 »

Offline vanibharutham

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Re: Additional Math Help HERE ONLY...!
« Reply #118 on: June 02, 2009, 10:42:10 pm »
sorry wasnt there....

ineedaz has got it :)
A genius is 1% intelligence, 99% effort.

Offline shan2391

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Re: Additional Math Help HERE ONLY...!
« Reply #119 on: June 02, 2009, 10:50:14 pm »
sorry wasnt there....

ineedaz has got it :)
ya but can u finsh off because i am totally lost