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,

Bob

]]>a) 2^5 is correct.

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

c) Just another Mississippi problem in disguise.

]]>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.

]]>You are welcome.

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

Bob

]]>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

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