Author Topic: Additional Math Help HERE ONLY...!  (Read 75478 times)

Offline sweetsh

  • Retired Adminstrator
  • SF V.I.P
  • ********
  • Posts: 7049
  • Reputation: 65535
  • Gender: Female
  • *Freedom
Additional Math Help HERE ONLY...!
« on: May 27, 2009, 05:05:08 pm »
Please post your doubts in Additional Math Subject so anyone can help.

Thank you
« Last Edit: May 30, 2009, 06:58:56 pm by sweetsh »

Offline shan2391

  • SF Immigrant
  • **
  • Posts: 103
  • Reputation: 3
Re: Additional Math Help HERE ONLY...!
« Reply #1 on: May 27, 2009, 05:14:53 pm »
permutation n combination

Offline sweetsh

  • Retired Adminstrator
  • SF V.I.P
  • ********
  • Posts: 7049
  • Reputation: 65535
  • Gender: Female
  • *Freedom
Re: Additional Math Help HERE ONLY...!
« Reply #2 on: May 27, 2009, 05:16:28 pm »
Post a question that would be easier to explain.

Offline Ghost Of Highbury

  • O_o_O lala!
  • SF Farseer
  • *******
  • Posts: 4096
  • Reputation: 41428
  • Gender: Male
  • Namaskaram!
Re: Additional Math Help HERE ONLY...!
« Reply #3 on: May 27, 2009, 05:17:57 pm »
hey sweetsh, have u taken additional math
i have it in my school
its fun

@shan
plz post a question in permutation and combination
so e can help
as very rightly said by sweetsh
divine intervention!

Offline sweetsh

  • Retired Adminstrator
  • SF V.I.P
  • ********
  • Posts: 7049
  • Reputation: 65535
  • Gender: Female
  • *Freedom
Re: Additional Math Help HERE ONLY...!
« Reply #4 on: May 27, 2009, 05:23:22 pm »
Yes I do.

Offline Anonymous

  • SF Immigrant
  • **
  • Posts: 81
  • Reputation: -6
Re: Additional Math Help HERE ONLY...!
« Reply #5 on: May 27, 2009, 05:28:00 pm »
Ask me any question and I'll pwn all you n00bs.

Offline Ghost Of Highbury

  • O_o_O lala!
  • SF Farseer
  • *******
  • Posts: 4096
  • Reputation: 41428
  • Gender: Male
  • Namaskaram!
Re: Additional Math Help HERE ONLY...!
« Reply #6 on: May 27, 2009, 06:20:16 pm »
@anonymous
i found this question in a guide and coul not solve it
hope u help me in this

Q. A train 150m long is running with a speed of 68mph. In whate time will it pass a man who is running at 8mph in the same direction which the train is going.

Q. A and B undertake to do a peice of work for $600. A alone can do it in 6 days while B alone can do it in 8 days. With the help of C
     they finish it in 3 days. Find the share of each?

i hope u help me.
thanks in advance
« Last Edit: May 27, 2009, 06:28:33 pm by ADi...M »
divine intervention!

Offline SGVaibhav

  • SF Farseer
  • *******
  • Posts: 3013
  • Reputation: 5737
  • Gender: Male
  • Bugatti Veyron 16.4
Re: Additional Math Help HERE ONLY...!
« Reply #7 on: May 27, 2009, 08:26:44 pm »
this looks impossible to the current maths i know                       (IGCSE MATHS)

Offline sweetsh

  • Retired Adminstrator
  • SF V.I.P
  • ********
  • Posts: 7049
  • Reputation: 65535
  • Gender: Female
  • *Freedom
Re: Additional Math Help HERE ONLY...!
« Reply #8 on: May 27, 2009, 08:28:37 pm »
It's similar to the math that you'll take in A Level.

Offline fREnZy

  • Newbie
  • *
  • Posts: 5
  • Reputation: 0
Re: Additional Math Help HERE ONLY...!
« Reply #9 on: May 28, 2009, 04:26:56 am »
Perms and Combs -
Whenever the order is necessary, use permutations. Whenever you just have to pick, in no order whatsoever, use combinations.

Offline divineobsidian

  • Newbie
  • *
  • Posts: 10
  • Reputation: 0
Re: Additional Math Help HERE ONLY...!
« Reply #10 on: May 28, 2009, 05:46:44 am »
can someone demonstrate a perm/comb and show logical steps?

like if there are 3 blue, 3 red, 4 greeen and you pick 5 how many perms/coms can there be?

thanks

Offline vanibharutham

  • SF Citizen
  • ***
  • Posts: 169
  • Reputation: 749
  • Gender: Male
Re: Additional Math Help HERE ONLY...!
« Reply #11 on: May 28, 2009, 09:52:31 am »
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
A genius is 1% intelligence, 99% effort.

Offline vanibharutham

  • SF Citizen
  • ***
  • Posts: 169
  • Reputation: 749
  • Gender: Male
Re: Additional Math Help HERE ONLY...!
« Reply #12 on: May 28, 2009, 11:43:29 am »
any more questions?

im happy to help...
A genius is 1% intelligence, 99% effort.

Offline sweetsh

  • Retired Adminstrator
  • SF V.I.P
  • ********
  • Posts: 7049
  • Reputation: 65535
  • Gender: Female
  • *Freedom
Re: Additional Math Help HERE ONLY...!
« Reply #13 on: May 28, 2009, 11:44:44 am »
Thank you for the information.

Offline Ghost Of Highbury

  • O_o_O lala!
  • SF Farseer
  • *******
  • Posts: 4096
  • Reputation: 41428
  • Gender: Male
  • Namaskaram!
Re: Additional Math Help HERE ONLY...!
« Reply #14 on: May 28, 2009, 01:55:25 pm »
Q. A train 150m long is running with a speed of 68mph. In whate time will it pass a man who is running at 8mph in the same direction which the train is going.

Q. A and B undertake to do a peice of work for $600. A alone can do it in 6 days while B alone can do it in 8 days. With the help of C
     they finish it in 3 days. Find the share of each??
hope u help..
divine intervention!