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#1 2012-09-07 16:41:57

genericname
Member
Registered: 2012-05-16
Posts: 52

Statistics/Probability(beginner): permutations and combinations

I'm having a hard time of understanding the meaning of those two and am also confused about which formula to use when given a problem related to those methods. What does it mean when it say things like "Order does not matter," or "No replacement"?

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#2 2012-09-07 19:46:56

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,469

Re: Statistics/Probability(beginner): permutations and combinations

hi genericname

Have a look at

http://www.mathsisfun.com/combinatorics … tions.html

Example:

How many three letter permutations can you make from the word MATH ?

MAT, MTA, AMT, ATM, TMA, TAM, MAH, ..........  24 answers.

How many three letter combinations can you make from the word MATH ?

MAT, MAH, MTH, ATH.  4 answers as order now doesn't matter.

Example of With replacement.

What is the probablity of drawing two aces from a pack of cards if the first card is replaced before the second is drawn.

P = 4/52 x 4/52

Example of Without replacement.

What is the probablity of drawing two aces from a pack of cards if the first card is NOT replaced before the second is drawn.

P = 4/52 x 3/51

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#3 2012-09-08 16:19:48

genericname
Member
Registered: 2012-05-16
Posts: 52

Re: Statistics/Probability(beginner): permutations and combinations

Thank you, Bob! That cleared things up.

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#4 2012-09-08 19:32:40

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,469

Re: Statistics/Probability(beginner): permutations and combinations

hi genericname

You are welcome.

Hope your studies go well.  But post again if you need more help.  smile

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#5 2012-09-09 14:44:17

genericname
Member
Registered: 2012-05-16
Posts: 52

Re: Statistics/Probability(beginner): permutations and combinations

This problem is really confusing me:

Consider five tosses of a coin. Assume that the results are arranged in order of toss.
a) What is the total number of possible outcomes to the five tosses?

b) In how many of those outcomes do all five of the coins show head?

c) In how many of those outcomes do exactly 3 of the coins show head?


For A I got 2^5 for an answer since there are 5 tosses and 2 choices each time(right?). I confused about what to do for B and C.

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#6 2012-09-09 15:37:22

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 89,048

Re: Statistics/Probability(beginner): permutations and combinations

Hi;

a) 2^5 is correct.

b)1 x 1 x 1 x 1 x 1 = 1

c) Just another Mississippi problem in disguise.


In mathematics, you don't understand things. You just get used to them.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

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#7 2012-09-09 17:05:14

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,469

Re: Statistics/Probability(beginner): permutations and combinations

hi genericname

It's not so many that you couldn't list them:

HHHTT
HHTHT
HHTTH
HTHHT
HTHTH
HTTHH
THHHT
THHTH
THTHH
TTHHH

So there's bobbym's 10 ways/

In general, you wouldn't want to do this though as the list could take ages to write out and you could easily miss some.

So how can you get this without ?

If the letters were all different, say H1 H2 H3 T1 T2, then I could re-arrange them in 5! ways.

But three of the letters are H's and so will look the same however I write them.  eg H1 T1 T2 H2 H3 would look the same as H2 T1 T2 H1 H3. 

So that 5! counts the same answer over and over.  How many times have I repeated the same answer because of the Hs ?

I can re-arrange H1 H2 H3 in 3! ways (H1 H2 H3;  H1 H3 H2; H2 H1 H3; H2 H3 H1; H3 H1 H2; and H3 H2 H1 )

So divide by 3! to allow for this.

Simily I've over counted the Ts ( T1 T2 would look the same as T2 T1) So I need to divide by 2! to allow for this.

Thus 5! / (3!x2!)

Hope that helps, smile

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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