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1. Morse code uses short pulses (dots) and long pulses (dashes). How many different symbols can be encoded if each symbol can use up to three pulses?
2. A drawer contains 6 Red socks, 8 green socks and 4 black socks.
a. In how many different ways can socks be taken from the drawer until a pair is found?.
b. What is the maximum number of socks which must be withdrawn before a pair is guaranteed?
c. What percentage of the solutions require exactly three socks to be withdrawn?
Hi Lucy10;
1)
You are saying up to 3 pulses so:
2 symbols for 1 pulse
2*2 = 4 symbols for 2 pulses
2*2*2 = 8 symbols for 3 pulses
2 + 4 + 8 = 14 different symbols.
2)
b) 3 is the maximum number you can withdraw before drawing a pair.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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No, 4 is the maximum!
Last edited by JaneFairfax (2010-02-06 02:04:24)
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Hi Jane;
Good to see ya. I reason like this, on the fourth one you have to have a pair, so 3 is the maximum before drawing a pair.
Oh no, she wants the number to draw a pair!!!!!! 4 is right. Read the question wrong. Sorry. I'm bleary eyed, over here. You take the rest of them, Jane. Thanks. Need some sleep.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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No problem. In fact I misread the first question myself. I thought the answer was 8 until you pointed out the teeny words up to at the end.
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2a.) How can they be drawn to pair?
I. Red Green Black Black (now you have 2 blacks)
II. Red Green Black Green (now you have 2 greens)
III. Red Green Black Red (now you have 2 reds)
IIII. Red Black Green Green
IIIII. Red Black Green Red
IIIIII. Red Black Green Black
IIIIIII. Green Black Red Red
IIIIIIII. Green Black Red Black
IIIIIIIII. Green Black Red Green
IIIIIIIIII. Green Red Black Black
XI. Green Red Black Green
XII. Green Red Black Red
XIII. Black Green Red Red
XIIII. Black Green Red Green
XIIIII. Black Green Red Black
XIIIIII. Black Red Green Red
XIIIIIII. Black Red Green Green
XIIIIIIII. Black Red Green Black
XIIIIIIIII. Red Red
XIIIIIIIIII. Green Green
XXI. Black Black
XXII. Black Red Red
XXIII. Red Black Black
XXIIII. Green Red Red
XXIIIII. Black Green Green
XXIIIIII. Red Green Green
XXIIIIIII. Green Black Black
XXIIIIIIII. Black Red Black
XXIIIIIIIII. Black Green Black
XXIIIIIIIIII. Green Red Green
XXXI. Green Black Green
XXXII. Red Green Red
XXXIII. Red Black Red
I probably missed something....
igloo myrtilles fourmis
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You can do it with numbers too if you want to.
1 = Green
2 = Black
3 = Red
Here are the ways to pull the socks out of the drawer:
11
22
33
121
131
212
232
313
323
122
133
211
233
311
322
1231
1232
1233
1321
1322
1323
2131
2132
2133
2311
2312
2313
3121
3122
3123
3211
3212
3213
So again 33 ways.
igloo myrtilles fourmis
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2c.) What percentage are 3 socks taken out of drawer, not 4 or 2, just 3 ?
I count 12 in the list of 33.
So answer is (12/33) times 100%.
That's 36.3636%
igloo myrtilles fourmis
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I stayed uplate until 3 am yesterday but could not figure it out the strategy using the permutation technique. Is there any way we can use nPr to solve it. Anyway you rescue my life !!!
Xoxoxoxo
If n = 5 and they are {A, B , C , D, E}
and r = 2, then
we get
5! / (5-2)! = 5 * 4 * 3 * 2 * 1 / (3 * 2 * 1) =
= 5 * 4 = 20
Let's list them to see if
this is similar to
the problem...
AB=#1
AC=#2
AD=#3
AE=#4
BA=#5
BC=#6
BD=#7
BE=#8
CA=#9
CB=#10
CD=#11
CE=#12
DA=#13
DB=#14
DC=#15
DE=#16
EA=#17
EB=#18
EC=#19
ED=#20
Hmmm.... take a look, and try to see if it can be used somehow...
igloo myrtilles fourmis
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Yes, it can be used, but it is not too easy.
If you look at my list of 33 numbers.
When the list has 4 digits, just look at
the first 3 digits, and that's 3! ways,
it's r=3 and n=3, so 3! / (3-3)! = 3! = 6.
And for the 3 digit ones, look at the
first 2 digits, still 3! or 6 because
r = 2 and n =3, so 3! / (3-2)! = 3! still. = 6.
igloo myrtilles fourmis
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For picking right out fast:
11
22
33
r = 1 for first digit, not looking at second digit.
and n = 3
3! / (3 - 1)! = 3 ways
So altogether 6 * 3 + 6 * 2 + 3 = 18 + 12 + 3 = 33 ways.
And I didn't explain the last digit multiplier.
that's the 3, 2 multipliers above in bold font.
igloo myrtilles fourmis
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Basically, if you pick out a pair, then the permutations have gone awry or bad.
Because the nPr permutation has no doubles in it.
So that's why I leave off the last digit, the last sock until later.
The permutations is everything up to and before picking the last sock.
igloo myrtilles fourmis
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Without the last digit or sock, if there are 2 digits,
then multiply nPr by 2 because each case has 2
different digits, and you want the pairs that go
with them.
Likewise for the 3 digit or sock case before getting a pair,
you have 3 unlike (different) digits, so you want
to multiply by 3 to get all the ways for those digits to
pair up.
And for the single digit, and then pick it right out,
just multiply by 1 to get the pair.
So in summary,
3P1*1 + 3P2*2 + 3P3*3 =
= 3 + 6*2 + 6*3 =
= 3 + 12 + 18 =
= 33
And the 12/33 * 100% is simply
the 3P2*2 / 33 times 100% (a form of 1).
igloo myrtilles fourmis
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Is there any way we can use nPr to solve it.
The formula is
Your first
socks must be different colours and you can pick them in ways. Your th sock must then be one of the colours you have already picked.Note that this formula is only for when you have at least as many socks of each colour as there are colours.
Last edited by JaneFairfax (2010-02-06 11:54:22)
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Sweet, that's the same thing as I said, just more concise!!
igloo myrtilles fourmis
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Thank you very much for your enthousiasm
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