Probability_q1 (1 Viewer)

[Richard Lee]

Hi. guys:

Try this question.

There are five letters, two A's, B, C & D. How many arrangements if two A's cannot be next to each other and B cannot be first.

Good luck!

 

[Archman]

4*4*3-2*(3+2+2+2) = 30,
or
3!*3+2*2*(2+1) = 30. which ever one you like

 

[Newbie]

can you explain it to me please mr archman

 

[Archman]

assume A is the first letter, the ways to choose the remaining is 3*3! (A cant be the second)
If A is not, lets look at the position of B C and D
since B cant come first, there are 2*2 ways of choosing the order of B,C and D, then theres (2+1) was of filling the A in each case.

 

[-=«MÄLÅÇhïtÊ»=-]

AABCD
Total Possiblities = 5!/2! = 60

B first and AA together
ie. *B*(A,A)CD
Possible perms = 3! = 6

So answer = 60 - 6 = 54

 

[McLake]

Originally posted by -=MLhtʻ=-
AABCD
Total Possiblities = 5!/2! = 60

B first and AA together
ie. *B*(A,A)CD
Possible perms = 3! = 6

So answer = 60 - 6 = 54

No, there are more possible perms that are of the form BAACD (2! postions for a 3! postions for AA block = 12, * 2 ways to place C and D = 24)

 

[...]

sorry, i dun do 4u, but i am very interested in this q...

would it be right if i work out unrestricted comb - the possible restricted comb??

5!-(4!*2!) = 120-50
=70

 

[-=«MÄLÅÇhïtÊ»=-]

Originally posted by McLake
No, there are more possible perms that are of the form BAACD (2! postions for a 3! postions for AA block = 12, * 2 ways to place C and D = 24)

eh?
3! accounts for all possible perms to put (a,a) , c , d
theres only 3 objects. Don't times it by 2! for the 2 a's coz they're identical. You actually times it be 2!/2!.

If you do it your way, there 3 positions for AA block, 2 positions for C and remanding 1 for D. 3*2 = 6

 

[McLake]

Originally posted by -=MLhtʻ=-
eh?
3! accounts for all possible perms to put (a,a) , c , d
theres only 3 objects. Don't times it by 2! for the 2 a's coz they're identical. You actually times it be 2!/2!.

If you do it your way, there 3 positions for AA block, 2 positions for C and remanding 1 for D. 3*2 = 6

But what about the C/D combos?

 

[-=«MÄLÅÇhïtÊ»=-]

already taken cared of by 3!
3! is 3P3. Like arranging 3 things in a line. i have 3 things block of AA, C, and D.

Minus it from 60 to get answer by using complements

 

[McLake]

Originally posted by -=MLhtʻ=-
already taken cared of by 3!
3! is 3P3. Like arranging 3 things in a line. i have 3 things block of AA, C, and D.

Minus it from 60 to get answer by using complements

Now i'm not sure ...

 

[-=«MÄLÅÇhïtÊ»=-]

ur rusty man =p
this time last yr, u wouldve gunned this question

 

[McLake]

Originally posted by -=MLhtʻ=-
ur rusty man =p
this time last yr, u wouldve gunned this question

Yeah, so would I ...

 

[ND]

Originally posted by -=MLhtʻ=-
AABCD
Total Possiblities = 5!/2! = 60

B first and AA together
ie. *B*(A,A)CD
Possible perms = 3! = 6

So answer = 60 - 6 = 54

But you didn't take into account when B is not first and the A's are together, and when B is first and the A's are apart.

 

[Newbie]

in these type of multiple condition sorta questions
should we always use 1 - complement thingy?

 

[sven0023]

A A B C D

5!/2! - 4! Ways of arranging if no A's next to each other = 36 ways.

Then there are 6 ways from these in which B is first
(4!/2! - 3!)

BACAD
BADAC
BACDA
BADCA
BCADA
BDACA

therefore, 36 - 6 = 30 ways.

 

[sven0023]

Similar to archmans;

Lets say A is first, there is 3!.3 ways of arranging the letters, then if C is first there is (4!/2!-3!) and if D is first then there is (4!/2!-3!) ways.

3!.3 + (4!/2!-3!).2 = 30 ways

 

[Richard Lee]

I think the answer is 30.

First thought:
How many arrangements for total:
the answer should be: 5P5/2!=60, since there are two A's.
Second thought:
How many arrangements for two A's togather:
the answer should be: 4P4=24
Third:
How many arrangements for the first position is B and two A's are not next to each other.
the answer should be: 4P4/2-3P3=6
Therefore, the answer that first position can't be B and Two A's can't be next to each other should be:
5P5/2!-4P4-(4p4/2!-3P3)=30

Therefore, I think the answer should be 30.

Go to "Probability-q2, if you want."

 

[-=«MÄLÅÇhïtÊ»=-]

Originally posted by ND
But you didn't take into account when B is not first and the A's are together, and when B is first and the A's are apart.

i didn't need to because the question says "and" not "or"

i c wat u guys have done now..
depends how you interpret the question i guess.

Edit: oh ye. sorry my bad. =p