Math Is Fun Forum

chetah

Member

Registered: 2008-02-15

Posts: 32

Discrete Maths

REPETITIVEE  are the given letters
1. How many different arrangements can there be?
2. How many arrangements with the EEs together?
3. How many arrangements with the two Is and the two Ts together?
4. How may arrangements with tR, P, V all next to one another?

need help with approach  to solve

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Discrete Maths

#4.  Can you clarify the tR thing?
  Do you mean  PVtR  VPtR   tRVP  tRPV  for starters
before you intermingle the other T and 6 vowels?
Oh, and PtRV and VtRP.

Last edited by John E. Franklin (2008-02-16 06:40:28)

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igloo myrtilles fourmis

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Discrete Maths

Chetah, could you please re-type your post more carefully and watch your spelling?

Don’t get me wrong – I am not trying to be rude – but the correct answers to your question depends on your stating it very carefully. For example, I am very sure the word is REPETITIVE, not REPETITIVEE. Are you sure it is REPETITIVEE?

We can only help you if you type your question carefully. If you don’t type correctly, we can at best give you answers that are wrong. So if you want us to help you with the correct answers, you must make sure you type carefully.

chetah

Member

Registered: 2008-02-15

Posts: 32

Re: Discrete Maths

REPETITIVE  are the given letters
1. How many different arrangements can there be?
2. How many arrangements with the EEs together?
3. How many arrangements with the two Is and the two Ts together?
4. How may arrangements with R, P, V all next to one another?

chetah

Member

Registered: 2008-02-15

Posts: 32

Re: Discrete Maths

Discrete Mathematics
The digits 1, 2, 3, 4, 5, 6, 7; consider strings of length 5.
Consider the number of such strings
1. The number of strings all digits are distinct
2. The number of strings where the digits are strictly increasing (the next digit > previous digit
3. How  many strings do not contain the string “1 2 3 4”

Asking for ideas in approaching this problem.

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Discrete Maths

1234567 #2.)
Here are the numbers you can leave out.
(The side-by-sides)
12, use other 5.
23, use other 5.
34, use other 5.
45
56
67
(the 6-aparts)
17
(the 5-aparts)
27
16
(the 4-aparts)
37
26
15
(the 3-aparts)
47
36
25
14
(the skip-aparts)
57
46
35
24
13
That's twenty-one of them.  7x6x5x4x3x2 / 2 / 5 / 4 / 3 / 2

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igloo myrtilles fourmis

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Discrete Maths

REPETITIVE #2.)
First alphabetize the letters:
EEE II P R TT V
Okay now we are ready to begin.
EEE II PR TT V   (#1)
EEE II PR TV T   (#2)
EEE II PR VT T   (#3)
EEE II PV RT T   (#4)
EEE II PV TR T
EEE II PV TT R
EEE II VP RT T
EEE II VP TR T
EEE II VP TT R
EEE II VR PT T  (#10)
EEE II VR TP T  (#11)
EEE II VR TT P
EEE IV IP RT T
EEE IV PI RT T
EEE IV PR IT T
EEE IV PR TI T
EEE IV PR TT I
EEE IV IR PT T
EEE IV IR TP T
EEE IV IR TT P (#20)
EEE IV IP TR T (#21)
EEE IV IP TT R
EEE IV PI RT T
EEE IV PI TR T
EEE IV PI TT R
EEE IV RI PT T
EEE IV RI TP T
EEE IV RI TT P
EEE IV TI PR T
EEE IV TI TR P (#30)
EEE IV TI RT P (#31)
EEE IV TI PT R
EEE IV TI TP R
EEE IV TI RP T
EEE IV IT PR T
EEE IV IT TR P
EEE IV IT RT P
EEE IV IT PT R
EEE IV IT TP R
EEE IV IT RP T (#40)
EEE IV TP IR T (#41)
EEE IV TP RT I
EEE IV TP RI T
EEE IV TP TI R
EEE IV TP TR I
EEE IV TP IT R
EEE IV PT RI T
EEE IV PT IR T
EEE IV PT TI R
EEE IV PT IT R (#50)
EEE IV PT TR I (#51)
EEE IV PT RT I
EEE II VT PR T
EEE II VT TR P
EEE II VT RT P
EEE II VT RP T
EEE II VT TP R
EEE II VT TR P
EEE II TV PR T
EEE II TV RP T (#60)
EEE II TV TR P (#61)
EEE II TV TP R
EEE II TV PT R
EEE II TV RT P
EEE II RV PT T
EEE II RV TP T
EEE II RV TT P
EEE II RP TT V
EEE II RP VT T
EEE II RP TV T (#70)
There are 290 more that start off
with EEE I...
Starting with EEE I...,
you get 24x5 + 24x4 + 24x3 + 24x2 + 24
because, well it's too hard for me
to say, and that's 24 x 15 or 360.
Next you intermingle the last I into
the rest like mathsy pointed out once.
And sometimes the I is in front of another
I and sometimes after another I, so those
look the same, so subtract one from the
equation coming up next...
Hold on...

This red part is WRONG, but I'll leave it here so you can see how I went haywire.
So that makes 5 x 360 more added onto
the 360, or 6 x 360 for 720 x 3, or
an olympic turning feat on snowboard,
2100 + 20 + 20 + 20 which is
2160 now.  But that's not the final
answer because the EEE is only at the
beginning.

Actually looking back, I should have said not 5 x 360 more,
but instead 2.5 x 360 more would work out right, and I
think it is because the I's got doubled-up in my thinking.

So now you intermingle or slide the
EEE's along all of these
IIPRTTV thingys.
So you can put the EEE in these positions:
EEE IIPRTTV   
I EEE IPRTTV 
II EEE PR TTV
IIP EEE R TTV 
IIPR EEE TTV
IIPRT EEE TV
IIPRTT EEE V
IIPRTTV EEE 
So that makes 8 ways you can
put the EEE's into the thousands of
other ones.
So 8 x 2160 is the answer for
#2 with the EEE's together.

2160
x    8
--------
17280 should be the answer!!  wrong answer

Last edited by John E. Franklin (2008-02-18 18:57:53)

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igloo myrtilles fourmis

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Discrete Maths

The last answer might be wrong!
I'm getting 6 mixes with PRV.
And then 6 x 10 = 60 for PRVTT's.
And then 21 x 60 for IITTPRV's = 1260 mixes.
And then 8 x 1260 for EEE's in it, which makes 10080 mixes.
Hhmm, I wonder which one is right?
...This one is right, it's 10080 mixes with EEE's together.

The problem I had with the letter I slipping into the
the other IPRTTV's, was that although I did see
that II is the same as II, I forgot that IxI was
different than IxI and IxxI was different from IxxI and
that IxxxI was different than IxxxI during intermingling because
the colored or bold "I" was already there, and the  other one was being
inserted.
So in IPRTTV's, if I is 1 below, but could be any of them in other cases,
So  I 1(skipI here if 1=I)2 I 3 I 4 I 5 I 6 I makes six cases, but
I neglected to think of somemore doubling up...
It's a little unclear, but I think I see it...
Well let's just list them...
I'll use I and J so it is not confusing.  "x" will be P, R, T, or V.
  I J x x x x x same as J I x x x x x
  J x I x x x x same as I x J x x x x
  I x x J x x x same as J x x I x x x
  OH I see, so 21 + 21 is 42.
21 is not 6 x 6, like my first mistake in other post.
3 x 7 is 21 and 3 x (2.5 + 1) = 21, not 6 x 6.
That's where I went wrong.  It's hard to explain,
but I'm just figuring it out both ways for myself.
Just ignore this because I can't explain it yet.
Gotta go to bed.

Last edited by John E. Franklin (2008-02-18 19:23:50)

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igloo myrtilles fourmis

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Discrete Maths

REPETITIVE  are the given letters
1. How many different arrangements can there be?
2. How many arrangements with the EEs together?
3. How many arrangements with the two Is and the two Ts together?
4. How may arrangements with R, P, V all next to one another?

1.
There are 10 letters, in which there are 3 E’s, 2 I’s and 2 T’s, so the total number of permutations is

2.
You mean with the EEE together? Treat the EEE as a single letter. Then there are 8 letters (EEE, R, P, T, I, T, I, V), in which there are 2 I’s and 2 T’s; hence the number of permutations is

3.
Use the same trick as before. Treat the II and the TT as single letters. Then there are 8 letters (II, TT, R, E, P, E, V, E), in which there are 3 E’s; hence the number of permutations is

4.
Same as before. Treat RPV as a single letter. Then there are 8 letters (RPV, E, E, T, I, T, I, E), in which there are 3 E’s, 2 I’s and 2 T’s. But now, there is something else: the letters RPV can arrange among themselves in 3! ways. Hence the number of permutations

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Discrete Maths

Discrete Mathematics
The digits 1, 2, 3, 4, 5, 6, 7; consider strings of length 5.
Consider the number of such strings
1. The number of strings all digits are distinct
2. The number of strings where the digits are strictly increasing (the next digit > previous digit
3. How  many strings do not contain the string “1 2 3 4”

Asking for ideas in approaching this problem.

1.

2.
For this I think it’s easiest just to list all the possibilities and count them:

34567

24567 23567 23467 23457 23456

14567 13567 13467 13457 13456 12567 12467 12457 12456 12367 12357 12356 12347 12346 12345

Hence the answer is 21.

3.
First count how many strings do contain 1234. Treat 1234 as a single digit; then there are 4 digits. The 1234 can occupy any of 4 places, and the other 3 places can be occupied by any number 1–7. Hence the number of such strings is 4×7[sup]3[/sup]. So the number of strings not containing 1234 is

chetah

Member

Registered: 2008-02-15

Posts: 32

Re: Discrete Maths

Discrete math
How can I determine the 13 position of the square root of 2 without using a recursive function starting from 2 and working my way up all the way up to the 13th position?