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s.sharma on September 20th, 2010 at 9:32 am

the answer should be 126.

method: 8 A's can be arranged in just one way. Now we have 9 places where 4 B's can go. 4 places out of 9 can be chosen in 126 ways. Therefore, the answer is 126.

Lalit on September 20th, 2010 at 10:53 am

we have to deduct the combination where two or three or four B's are standing next to each other and filling gaps between A's or B's is not right as that will be Derangement (12 places being fixed in number i.e. have to ignore extreme left/right places). Best way is to tie the redundant B's in a string, so BBBB as "X" or BB BB as "X X" and BBB B as "X B". In this way, we deal with straight formula arranging X-8A's, XX8A's,XB8A's..all of which are different. And lastly..we know in X, its only way arrangement.
12!/8!4! - 9!/8! - 10!/8!2! - 9!/8!

Sunil on September 20th, 2010 at 11:45 am

Hi,

The answer is 126.

The simplest way to approach these questions is, first of all arrange those things on which there is no constraint in the problem.

In this question, there is no constraint on the arrangement of a.

Number of ways of arranging the 8 identical a's = 1

After the arrangement of a's is over, it will look loke this

_a_a_a_a_a_a_a_a_

So, we can clearly see that, 9 spaces are created in the arrangement of 8 a's. So to arrange 4 b's, we can select 4 spaces out of these 9 spaces.

So, number of ways to arrange 4 identical b's in 9 places = 9C4*1

So, total number of ways of arranging a's and b's together = (No of ways of arranging a) * (No of ways of arranging b)

1*9C4 = 126

Sunil on September 20th, 2010 at 12:04 pm

The simplest way to approach these questions is to arrange first those things on which no constraint is given in the problem.

In this question, there is no constraint on the arrangement of a's, so arrange a's first.

No of ways of arranging the 8 identical a's = 1

Now, after the arrangement of a's, 9 spaces are created among them:

XaXaXaXaXaXaXaXaX

As we can clearly see that 9 spaces are created after the arrangement of 8 a's. Now arrange 4 b's in these 9 spaces so that the constraint of no 2 b's together is not violated:

No of ways of arranging 4 identical b's in 9 spaces = 9C4*1

Total number of ways of arranging all a's and b's = (No of ways of arranging the 8 identical a's) * (No of ways of arranging 4 identical b's) = 1*9C4 = 126

neha on September 20th, 2010 at 2:42 pm

all of u took gr8 care in not taking two Bs together but no body paid heed to the arrangements wen two or more A can come together
so it shud go the following way
when no As ttogether like XAXAXAXAXAXAXAXAX
9 places for B = 9C4=126
when 2 As together consider two A as one entity like XAXAXAXAXAXAXAAX we have only 8 places for B now =8C4= 70
similarly when 3As together then 7C4= 35
when 4 As together then 6C4= 15
when 5 as together then 5C4 = 5
and lastly when 6 as together then 4C4=1
summing up all gives 252

Nik on September 20th, 2010 at 11:37 pm

9 empty spots to be filled by 4 Bs.
Answer: 9C4

BG on September 21st, 2010 at 12:34 am

The answer is 126 ways.

_A_A_A_A_A_A_A_A_

You have 9 possible places to put the 4 B´s, therefore 9C4 = 126

Phiroz on September 21st, 2010 at 4:54 am

9c4 = 126

Lalit on September 21st, 2010 at 10:05 am

Still not getting, why in 8As.. two A's can not come together. viz in lieu of _A_A_A_A_A_A_A_A_ .... I still can have AAA_AA_A_A_A. Basically my doubt was left-right of a B is there, For A its not.