Probability-q2 (1 Viewer)

[Richard Lee]

There are two six letters: two A's, two B's, one C and one D. How many arrangements if two A's and two B's cannot be next to each other?

Good luck!

 

[ND]

I'll have a go but i don't know if it's correct:

Total number of arrangements: 6!/2!*2!
Arrangements with A's together: 5!/2!
Arrangements with B's together: 5!/2!
Arrangements with both A's and B's together: 4!

Total number of arrangements without A's or B's together = Total no. of arrangements - A's together - B's together + both A's and B's together = 6!/2!*2! - 5! + 4!

 

[sven0023]

hmm, i'm in yr11 so this is probably wrong, ill give it a go neway;

A A B B C D

All together : 6!/2!2! = 180ways
TwoA's and TwoB's together : 4! = 24ways
OnlyA's together : 5!/2! - 4! = 36ways
OnlyB's together = 36ways

Therefore, 180 - 24 - 36 - 36 = 84ways.

 

[sven0023]

Also could anyone help me with this question?

Condsider the letters A A A B B B C C C

If a 5 letter word is to be made out of these letters,
how many ways can the letters be arranged if;

i) No restrictions
ii) A's cannot be next to each other
iii) A's cannot touch C's

My teacher gave us a sheet with this qn on... im confident in (i) and (ii) yet I'm stuck on (iii)... any help?

 

[freaking_out]

Originally posted by sven0023
Also could anyone help me with this question?



My teacher gave us a sheet with this qn on... im confident in (i) and (ii) yet I'm stuck on (iii)... any help?

lol...can u remind me how to do (i)?

 

[Constip8edSkunk]

iii)
there are 4 cases of XXYYY where X,Y=/= A,C
there are 2 cases [2a2b1c or 2c2b1a]: the number of arrangements for each: _cb_(a_a)_, where a_a and c can swap places, (2*2*2) + abcba (1)
2 cases of either 3a or 3c with 1 b, each with 2 possible arrangements
1 case of 3b1a1c, with 2*4C2 (a and c interchangeable, 4 spaces choose 2 for b)arrangements
the remaining case 2C2A1B has 2 arrangements
(5!/3!2!) *4 + (2*2*2+1) *2 + 2*2 +2*4C2+2 = 76

 

[MyLuv]

I dont really understand the question,anyway here is my answer: ( I guess the question want U have both A 'n C in the five,if not U just add all the case where there's 1 of A or C )

If A cant touch C ,B must be in the five:
+ if 1 B: obviously there're 4 ways : AAABC,CCCBA,AABCC,CCBAA
+if 2 B: U can choose 2A ,1C (or 2C,1A) ,'n for each case:
. No. of ways A,C touching: 3!*3!/2!*2!=9
. Toal No.= 5!/2!*2!=30
---> No. of way if there're 2 B = 2*(30-9)=42
+if 3B : U can choose only 3 B,1A,1C
.No. of ways A,C touching : 2*4!/3!=8
.Total No.= 5!/3!=20
--> No.of ways if there 're 3B =20-8=12
>>> Total No. for A,C not touching=4+42+12=58
ps: I didnt count case there're only B 'n A or B 'n C

 

[freaking_out]

hey, so how do i do part (i)?

 

[MyLuv]

Hehe,for part 1: there're 3 casesnot sure)
+ 3 same,2 same (eg:3A,2B): 6* 5!/3!*2!=60
+ 3 same,2 diff (eg:3A,1B,1C) : 3*5!/3!=60
+ 2 same,2same,1 other (eg:2A,2B,1C) : 3*5!/2!*2!=90
--> Total=60+60+90=210

 

[Constip8edSkunk]

i think for the third part there are less than 42 arrangements for 2 Bs, heres what i thought: 2 possible cases: 2a, 2b 1c or 1a 2b 2c. take the first case, the As and the Cs must be separate and there may be a spot between the As:
<c>_(A_A)_BC_ </c>
there are 4 places to place the B and times this by 2 as (A_A) can swap with C. there is an extra case when the C is between the 2 As. .'. there are 9 arrangements for this case ans 18 arrangements in total for the 2Bs.

 

[MyLuv]

Originally posted by Constip8edSkunk
. there is an extra case when the C is between the 2 As. .'. there are 9 arrangements for this case ans 18 arrangements in total for the 2Bs.

---> I dont really understand what U meant(coz of mine bad English ). Anyway,if C is between As then they have touched<--???
Explain for 2B in mine solution: let AAC be in a group(for touching) the others r 2 B --> no.of way is 3!/2! 'n in AAC there 's 3!/2! no.of arrangement.--> No. of way A & C touching in case of 2B,2A,C is 3!*3!/2!*2!

 

[Hotdog1]

Oh God I feel stupid. I don't understand either of you, so i just counted the possibilities, there aren't that many.

AABBC
AABCB
ABABC
BAABC
BCBAA
CBAAB
CBABA
CBBAA

I can only find 8 arrangements with 2 B's in the word. Hmmmmmm

 

[SoFTuaRiaL]

1. 2 ways of arranging c,d
2. As can be arranged in the 3 spaces in c(3,2) ways
3. Bs can be arranged in the 5 spaces in c(5,2) ways

total = 2*6*10 = 60

edit: this is the answer to the first original uestion

 

[MyLuv]

Oh,I forgot to count AC case in my solution ;anyway I redo it 'n got this:
In 1B case there r 6 way
In 3B case there r 12 way
In 2Bs case there r 3 case:
C is the 1st--->(CB) is 1st ,there're 3!/2!=3
C is the last---> (BC) is last, there're 3!/2!=3
C in the middle--> (BCB) is a term, there're 3!/2!=3
Similar for 2C ,2B ,1A-->Total= 2*9=18 ways for 2 Bs
--> Total No. of way= 6+18+12=36
If u count case there r only B 'n A (or C) then total No. is 36+4*5!/3!*2!=76

 

[SoFTuaRiaL]

is this the soln to the first or the 2nd q?