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Dumbarse

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1) How many different six letter words can be formed from the letters of the word AUSTRALIA ??

2) How many different sums of money can be made up from six $20 notes, two $5 notes, four $2 coins, seven $1 coins, three 50c coins, and one 20c coin ??
 

BlackJack

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2) 147 * 4 (including $0 as a sum)
=588
explanation:
can make 0,20c,50c,70c regardless of amount ( * 4 bit)
total sum of money = $146.70
we can make every dollar value in between (think about this...)
possible dollar values= 147.

1) I don't know... I split it up into Ways( 0 A) + ways(1 A) + ways(2 A) + ways( 3A)

A, A, A, U, S, T, R, L, I

w(0A) = 6P6 = 720
w(1A) = 6P5 * 6 = 4,320
w(2A) = 6P4 * 6*5/2 = 5,400
w(3A) = 6P3 * 6*5*4/6 = 2,400

sum = 12,840
It could be wrong, but I have no idea how to check it.
 

spice girl

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Originally posted by Dumbarse
1) How many different six letter words can be formed from the letters of the word AUSTRALIA ??

2) How many different sums of money can be made up from six $20 notes, two $5 notes, four $2 coins, seven $1 coins, three 50c coins, and one 20c coin ??
1) There are 3 'A's in australia, 7 different characters altogether.

So this is like the other problem that fuelled the huge argument in the FREQUENT thread. The solution is to break it up into cases:

No 'A's: 6 different letters, all must be included: 6!

1 'A': 7 different letters, 'A' must be included:
'A' can be in any of the six, then for the remaining 5 slots there are 6, 5, ..., 2 options remaining: so it's 6*6*5*4*3*2

2 'A's: Choose two slots for the two 'A's, then 6, 5, ..., 3 options remaining for the 4 other slots: (6C2)*6*5*4*3

3 'A's: Same idea: (6C3)*6*5*4

Altogether: 12840


2) The strategy here is to determine how many different amounts between 0 to $20, you can make (i.e look at all the possible remainders when you divide by 20). Since this can be repeated when you add a $20 note, for all the way up to $120. Then you must determine how many different options above $120 you can make:

One 20c coin, more than enough 50c coins: can only make $x.00, $x.20, $x.50, $x.70 (4 choices)

More than enough $1 and $2 coins: unit figure can be anything from 0 to 9 (10 choices)

2 $5 notes: can make anything from 0 to 19 (mod 20) (multiply choice by 2)

So there are 4 * 10 * 2 possible remainders, and since you can have 0, 1, 2, ..., 5, you multiply by 6: total 480.

Now, you've got to determine how many different options above $120 you can make.

Now whether you have 5 or 6 $20 notes, it won't give you any more options, so lets assume 6

So must determine how many options you can make WITHOUT the twenties notes:

From less than $20: 80 options (determined above)
More than $20: can make up to $25 using gold coins. (6*4 = 24)
$26 or over: can make $26.00, $26.20, $26.50, $26.70 (=4)
So 108 options above $120

Total options (including zero sum as an option) is 588
Excluding zero as an option: 587

Edit: which is the same as blackjack, which is cool
 
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Dumbarse

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i like spice girl's answer for the first one
and blackjacks answer for the second one is heaps good, nice logic there, the way they did it in the textbook was long and stoopid

thanx people
:wave:
 

Weisy

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Another question

Calculate the probability of getting a straight (five consecutive cards, any suit) in a game of poker

First: I've never played poker; can a straight cross the end of the number values, ie. jack, queen, king, ace, two? if so, I didn't account for them in my calculations.

Second: I got 9216 out of a possible 2598960 (52C5) combinations, but I'm not sure if and I don't have the answers.

any thoughts?
 

Weisy

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Another question

Calculate the probability of getting a straight (five consecutive cards, any suit) in a game of poker

First: I've never played poker; can a straight cross the end of the number values, ie. jack, queen, king, ace, two? if so, I didn't account for them in my calculations.

Second: I got 9216 out of a possible 2598960 (52C5) combinations, but I'm not sure I did it the right way and don't have the answers.

any thoughts?
 

wogboy

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I think that a straight in poker means where you have 5 consecutive cards including the royalties (King, Queen, Jack, Ace), but the rule goes that if you get an Ace it can only be the highest or lowest card you have in that straight. That means that the possible numbers you can have in a straight are (regardless of the suits of the cards):

A,2,3,4,5
2,3,4,5,6
3,4,5,6,7
4,5,6,7,8
5,6,7,8,9
6,7,8,9,10
7,8,9,10,J
8,9,10,J,Q
9,10,J,Q,K
10,J,Q,K,A

= 10 different number arrangements.

Because there are 4 different suits of the same numbers, and there are 5 cards in a straight,

the total no of ways that a straight can occur = 10 * 4^5

the total possible no of any 5 cards given at random = 52c5

therefore, the probability of a straight is = 10 * 4^5 / 52c5

= 10240/2598960
= 1024/259896
 

BlackJack

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*smiles* Somebody called?

Can't see anything wrong w/ wogboy's... how did you get yours?

fraction simplified: 128 / 32487
---
What about the chances of getting a flush?:D
 

Weisy

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lol

no, that's fine. my answer was only off because I didn't know that you could have the extra 10, J, Q, K, A arrangement

So I had only 9 different number arrangements, and then I multiplied by 4^5/52C5 to get 9216/2598960.

A lack of knowledge of poker, rather than a mathematical deficiency, although I'm sure I have both somehow. :p

Now, I'll try the flush...five cards all of the same suit (right?)

In one suit there are 13c5 ways of getting a flush. In four suits, there would be 13c5 x 4 ways. So total probability is 13c5x4/52c5.

= 5148/2598960

= 33/16660

=1.98 x10^3

any more?
 

Dumbarse

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(u will never get asked a card question in the HSC people!)

2) How many different sums of money can be made up from six $20 notes, two $5 notes, four $2 coins, seven $1 coins, three 50c coins, and one 20c coin ??

the answer is not 588 peoples,
this is how the book did it which i cannot understand at all

answer:
using (p+1) (q+1) (r+1)... -1

number of sums of money

= (6+1) (2+1) (4+1) (7+1) (3+1) (1+1) -1

= 7.3.5.8.4.2 -1

= 6719

please explain....
 

BlackJack

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Hmmm. :/ They don't care if the final amount's the same...
the result comes from the choices of using how many of each note/coin.
For the $20, 6 meaning using 1 to 6 $20 notes in the combination, +1 meaning the possibilty of using no $20 nots. therefore 7 possible ways of choosing the $20... Ditto for other currencies. The -1 at the end takes out the final usm of $0...

I'm still cross about 'DIFFERENT sums' though...
 

BlackJack

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Originally posted by Raser
12C1 * 4C3 * 11C1 * 4C2 / 52C5 = 66/54145

that anywhere near right?
If you're answering the full house...
there are 13 sorts of cards (A-K), so not quite... 13C1 * 4C3 * 12C1 * 4C2 / 52C5
6/4165
 

Weisy

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ah Blackjack, sorry i missed the fun.

to make up for it, I did all of them. (it was either that or banked tracks)

Order of winning hands (power)

royal flush/routine =1/649740
straight flush = 3/216580
four of a kind = 13/49980
full house = 6/4165
flush = 33/16660
straight = 128/3248
three of a kind = 13/490
two pair = 292032/2598960 (can't be bothered simplifying anymore)
two of a kind = 1723800/2598960

hmmm...I wonder how many are wrong...

eat your heart out. :)

:)
 

BlackJack

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Actally, in Asian countries internal organs are a delicacy... animal ones of course...
My version of probabilities, I've excluded straights from being flushes, etc....:
Royal flush: 1/649740
Straight flush: 3/216580
four of a kind: 1/4165 (13* 48 ways)
full house: 6/4165
flush: 1277/649740 ( (13C5-10) * 4)
straight: 5/1274 ( (4^5 - 4) * 10)
three of a kind: 88/4165 (4C3*13 * 48*44 / 2)
two pairs: 396/4165 (4C2*13 * 4C2*12 * 44)
one pair: 352/833 (4C2*13 * 48*44*40/3!)
In other words, the relative chances:
RF: 1
SF: 9 (10 counting RF)
FK: 156
FH: 936
FL: 1,277 (1287 counting SF)
ST: 2,550 (2560 counting SF)
3K: 13,728
2P: 61,776
1P: 274,560
You'll get something around 54.6361621...% of the time... 1419972...
I wonder how many are wrong as well... edit: there is probably something wrong... think about
(4^5-4)(13C5-10) = 1,302,549... around 50.1177394...% to get nothing...
 
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