Fosweb
I could be your Doctor...
How many ways are there of arranging AABBCCDD so that no two of the same letters are next to each other?
How do you do this please?
How do you do this please?
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More probability. (Simple?) (1 Viewer)
- Thread starter Fosweb
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Constip8edSkunk
Joga Bonito
4!*4=96 ?
N
ND
Guest
I got the same:
Put 4 single letters down first (there are 4! ways of doing this), then there is 4 ways to slot the remaining letters in so that no 2 of the same letters are touching.
Put 4 single letters down first (there are 4! ways of doing this), then there is 4 ways to slot the remaining letters in so that no 2 of the same letters are touching.
Fosweb
I could be your Doctor...
the answer is 864... but how do you get this?
what i have done: (And there must be something easier?)
Total ways of arranging the 8 letters:
8!/(2!2!2!2!) = 2520
Then:
2520 - 4PairsTogether - 3PairsTogether - 2PairsTogether - 1PairTogether
= 2520 - 4C1(4!) - 4C2(5!/2!) - 4C3(6!/(2!2!) - 4C1(7!/(2!2!2!))
But this doesnt take into account that when you have 3 or 2 or 1 pair together, then remaining numbers cant occur in pairs.
Confused even more now... hmm.
what i have done: (And there must be something easier?)
Total ways of arranging the 8 letters:
8!/(2!2!2!2!) = 2520
Then:
2520 - 4PairsTogether - 3PairsTogether - 2PairsTogether - 1PairTogether
= 2520 - 4C1(4!) - 4C2(5!/2!) - 4C3(6!/(2!2!) - 4C1(7!/(2!2!2!))
But this doesnt take into account that when you have 3 or 2 or 1 pair together, then remaining numbers cant occur in pairs.
Confused even more now... hmm.
Last edited:
Archman
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since all letters are interchangeable, assume the first letter is A, then the second cant be A , so let it be B
AB
now there r 2 cases
ABA or ABC
then that spread to
ABAC, ABAB or
ABCA, ABCD, ABCB
for ABAB there r 1 way of putting C and D after it
for ABCD there r 3*3*2*1 = 18 ways
ABAC can expand to ABACD, ABACB
for ABACD there r 2*2*1 = 4 ways
for ABACB there r only one way of filling the rest: DCD
ABCA can expand to ABCAC, ABCAB and ABCAD
ABCAC: one way DBD
ABCAB: one way DCD
ABCAD: 2*2*1 = 4 ways
ABCB is similar to ABCA there should be 6 ways
so there are 1+18+4+1+1+1+4 + 6= 36 ways
36 * 4! = 864 yay
this is really the dum way where your count every case.
p.s i can times it by 4! because in all the 36 ways, if u remove the second A,B,C and D
u get ABCD
AB
now there r 2 cases
ABA or ABC
then that spread to
ABAC, ABAB or
ABCA, ABCD, ABCB
for ABAB there r 1 way of putting C and D after it
for ABCD there r 3*3*2*1 = 18 ways
ABAC can expand to ABACD, ABACB
for ABACD there r 2*2*1 = 4 ways
for ABACB there r only one way of filling the rest: DCD
ABCA can expand to ABCAC, ABCAB and ABCAD
ABCAC: one way DBD
ABCAB: one way DCD
ABCAD: 2*2*1 = 4 ways
ABCB is similar to ABCA there should be 6 ways
so there are 1+18+4+1+1+1+4 + 6= 36 ways
36 * 4! = 864 yay
this is really the dum way where your count every case.
p.s i can times it by 4! because in all the 36 ways, if u remove the second A,B,C and D
u get ABCD
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Constip8edSkunk
Joga Bonito
hahaha i couldnt sleep last night so i ended up doing it that way... are there a more simpler way tho? oh yeah ND, our original solution is screwed cuz it only assumes 1 letter between the sets, eg ABA...s are excluded.
Consider "formula" for " exactly one same letter pairing":
(1) 4C1 7!/2!2!2!
This is obviously wrong as it already equals 2520=8!/2!2!2!2! ie. total number of ways.
There is massive overcounting in (1). In fact the corresponding formula for "exactly" two pairings
(2) 4C2 6!/2!2! is double-counted in (1) because two pairings AA,BB are counted twice. Similarly, corresponding formulas for 3 and 4 pairings eg. 4C3 5!/2! and 4C4 4! are triple and quadruple counted in (1).
Unfortunately, the following would be a "knee-jerk" overcorrection
4C1 7!/2!2!2!-4C2 6/2!/2!- 2*4C3 5!/2!- 3* 4C4 4!
The reason for this is that the correction factor 4C2 6/2!/2! already goes some way in correcting the triple-counted 4C3 5!/2!, and so on.
The following adjustments give the number of arrangements with at least two same letters next to each other.
4C1 7!/2!2!2!-4C2 6/2!/2!+4C3 5!/2!- 4C4 4! =1656
Just a further point of clarification, the last correction term
- 4C4 4! was obtained by the following reasoning,
it not only was included in 4C1 7!/2!2!2! but 4 times as many had been included, -4C2 6/2!/2! subtracted 6 times as many, +4C3 5!/2! added 4 times as many, finally the last correction term brings it down to just one.
Thus number of arrangements with no two same letters next to each other is 2520-1656=864.
A hard one.
(1) 4C1 7!/2!2!2!
This is obviously wrong as it already equals 2520=8!/2!2!2!2! ie. total number of ways.
There is massive overcounting in (1). In fact the corresponding formula for "exactly" two pairings
(2) 4C2 6!/2!2! is double-counted in (1) because two pairings AA,BB are counted twice. Similarly, corresponding formulas for 3 and 4 pairings eg. 4C3 5!/2! and 4C4 4! are triple and quadruple counted in (1).
Unfortunately, the following would be a "knee-jerk" overcorrection
4C1 7!/2!2!2!-4C2 6/2!/2!- 2*4C3 5!/2!- 3* 4C4 4!
The reason for this is that the correction factor 4C2 6/2!/2! already goes some way in correcting the triple-counted 4C3 5!/2!, and so on.
The following adjustments give the number of arrangements with at least two same letters next to each other.
4C1 7!/2!2!2!-4C2 6/2!/2!+4C3 5!/2!- 4C4 4! =1656
Just a further point of clarification, the last correction term
- 4C4 4! was obtained by the following reasoning,
it not only was included in 4C1 7!/2!2!2! but 4 times as many had been included, -4C2 6/2!/2! subtracted 6 times as many, +4C3 5!/2! added 4 times as many, finally the last correction term brings it down to just one.
Thus number of arrangements with no two same letters next to each other is 2520-1656=864.
A hard one.
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Fosweb
I could be your Doctor...
OLDMAN: thanks. I did say above that i knew this was wrong (with 3 or 2 or 1 exactly pairs) for this reason, i just couldnt work out how to get rid of the other pairs that could occur also.
However: it this still the easiest way to do this? (I mean: taking the probability of all the 'not' cases combined from 1)
Or is there some other way to do it?
However: it this still the easiest way to do this? (I mean: taking the probability of all the 'not' cases combined from 1)
Or is there some other way to do it?
_________________________________________________
Quote Fosweb:
OLDMAN: thanks. I did say above that i knew this was wrong (with 3 or 2 or 1 exactly pairs) for this reason, i just couldnt work out how to get rid of the other pairs that could occur also.
However: it this still the easiest way to do this? (I mean: taking the probability of all the 'not' cases combined from 1)
Or is there some other way to do it?
__________________________________________________
Yes, your initial analysis gave me a start.
Would be interested in a more elegant solution. I am not sure whether even brilliant students could reproduce the reasoning behind under exam conditions. I myself required a good night's rest and early morning mindclear for this one.
Quote Fosweb:
OLDMAN: thanks. I did say above that i knew this was wrong (with 3 or 2 or 1 exactly pairs) for this reason, i just couldnt work out how to get rid of the other pairs that could occur also.
However: it this still the easiest way to do this? (I mean: taking the probability of all the 'not' cases combined from 1)
Or is there some other way to do it?
__________________________________________________
Yes, your initial analysis gave me a start.
Would be interested in a more elegant solution. I am not sure whether even brilliant students could reproduce the reasoning behind under exam conditions. I myself required a good night's rest and early morning mindclear for this one.
turtle_2468
Member
Just out of interest, this appeared in an actuarial quiz in UNSW around 2 weeks ago. Needless to say, no one got it.. 
This looks very much like an Actuarial probability question... and considerin' I'm doin' Combinatorial Probability, I should have a go... alas, some other time.Originally posted by turtle_2468
Just out of interest, this appeared in an actuarial quiz in UNSW around 2 weeks ago. Needless to say, no one got it..
As an actuarial student you'd be doing it using the Principle of Inclusion-Exclusion (which is what the method OLDMAN is using is called in more formal terms).
My comment on that UNSW Actuarial Quiz - any student who didn't pass it (at the least) should be disappointed (as much for morale as for its actual standard). For those who aren't UNSW actuarial students, examples of the questions include:
- find the coefficient of x^99 in (x-1)(x-2)...(x-100) (21 marks)
- write your tutor's name in the top left-hand corner of the answer book (5 marks)
In terms of the HSC, the only method I can see to do it is a case-bash, which is again excessively messy:
Assume the order the letters appear is A,B,C,D (so we multiply our answer by 4! at the end).
Then we have ABCDD (no D can appear before the first C)
There are 4 places for the second A (after the first B, since it can't be next to the existing A).
There are now 4 places for the second B (after the letter after the first B, since it can't be before the first A or next to the existing B)
Then we case-bash on the letters between the two D's:
(1) If no letters are presently there, the second C must go there. This is 9 of the existing cases, and we have 9 ways.
(2) If only an A is presently there, then there are 4 places for the second C to go. This takes care of 2 cases, giving 8 ways.
(3) If both the second A and second B are there, then the second C can go into any of 4 places, taking care of 2 more cases to give 8 ways again.
(4) If only a B is presently there, then in 2 cases the second A will be after the first C, giving 4 ways in each case, and in the third the A will be between the first B and first C (i.e. ABACDBD), so that there will be just 3 places to put the second C down. This gives 11 ways.
Hence we have 9+8+8+11 = 36 ways, and then 864 in total.
My comment on that UNSW Actuarial Quiz - any student who didn't pass it (at the least) should be disappointed (as much for morale as for its actual standard). For those who aren't UNSW actuarial students, examples of the questions include:
- find the coefficient of x^99 in (x-1)(x-2)...(x-100) (21 marks)
- write your tutor's name in the top left-hand corner of the answer book (5 marks)
In terms of the HSC, the only method I can see to do it is a case-bash, which is again excessively messy:
Assume the order the letters appear is A,B,C,D (so we multiply our answer by 4! at the end).
Then we have ABCDD (no D can appear before the first C)
There are 4 places for the second A (after the first B, since it can't be next to the existing A).
There are now 4 places for the second B (after the letter after the first B, since it can't be before the first A or next to the existing B)
Then we case-bash on the letters between the two D's:
(1) If no letters are presently there, the second C must go there. This is 9 of the existing cases, and we have 9 ways.
(2) If only an A is presently there, then there are 4 places for the second C to go. This takes care of 2 cases, giving 8 ways.
(3) If both the second A and second B are there, then the second C can go into any of 4 places, taking care of 2 more cases to give 8 ways again.
(4) If only a B is presently there, then in 2 cases the second A will be after the first C, giving 4 ways in each case, and in the third the A will be between the first B and first C (i.e. ABACDBD), so that there will be just 3 places to put the second C down. This gives 11 ways.
Hence we have 9+8+8+11 = 36 ways, and then 864 in total.
Richard Lee
Member
Hi, there.
Try this calculation:
4P3(5P5/2-4P4)=864
Try this calculation:
4P3(5P5/2-4P4)=864
Constip8edSkunk
Joga Bonito
and the reason behind this is?
Rahul
Dead Member
good luck in tryng to get that....all his posts seem to be 2 or 3 lines or 'working' at most. unlike other posts above....
Bannanafish
Member
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richard??
Last edited:
Richard Lee
Member
It's hard to explain!
Sorry about!
Go to the thread : Probability_q1.
I will post the couple of questions, like this. If you can follow it, you will know how to work out this question!
Good luck!
Richard Lee
Member
Go to the thread: Probability_q1, plz!
You will find the answer by youself.
Good luck!
Richard Lee
Member
Good guy!Originally posted by Lazarus
Uhh... surely a good tutor would be able to explain difficult concepts... ?
I think tutors would become rather redundant if solutions could be gleaned by simply looking at questions...
I think you are right. You can find TUTOR everywhere. But there are a few good tutors. So, don't worry about the redundant since the bad tutor will disappear soon. $60 per? It must be a good tutor. Trust me!
I couldn't explain since I need more room and time. But as a good tutor, I know how to let my friends to understand the qeustion and method, not just the answer. Actually, I got a lot of hard question and I ask my friends to do them. But I don't really know the answer. Since, I don't care. I just care the process or method or idea.
Anyway, Go to the thread "Probability-q1" , and follow the questions, you will know how to work it out by yourself.
I am sure, if you can follow it, you can improve yourself a lot!
Don't say anything, plz. Just try. When you finish the question, come back and talk to me, again!
Thanx!