Number of ways in which 5 A’ and 6 B’s can be arranged in a row which

Question

Number of ways in which 5 A’ and 6 B’s can be arranged in a row which reads the same backwards and forwards, is

(A) 6 
(B) 8
(C) 10 
(D) 12
(E) 14

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sujoykrdatta · Answer

There are 5 As - so 2 on either side and one A in the middle
There are 6 Bs - so 3 on either side

So there are 5 letters on the left and 5 on the right with one A in the middle

Number of arrangements on the left side 
= 5!/(2! x 3!) = 10

We just need to check the arrangements on one side since each arrangement on the left must mirror the same arrangement on the right.

Hence, there are 10 ways. 
Answer C

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GMATinsight · Answer

- We have to arrange the 11 letters (5 A’ and 6 B’s)

- The arrangement must be palindrome which read same backwards and forwards

- For the result of 11 letters to be palindrome, one letter must be pivoted at 6th place  - - - - -  -  - - - - -

- The pivoted letter MUST be an A because there are 5 A's so we can't divide them into two equal parts for result to be palindrome
 - - - - -  A  - - - - -

- Now our objective is only to arrange the 5 letters to the left of middle letter as an arrangement to the left will be mirrored as arrangemenbt to teh right of pivoted letter
  - - - - -   A  - - - - -

- The  5 letter to the left , must be 2 A's and 3 B's

- Arrangements of  2 A's and 3 B's  can be counted in 5C2 ways because we only need to identify the two places for 2 A's out of 5 empty places.
- No arrangement need to be taken into account because all A's are identical and all B's are identical

So desired outcomes =  ^5C_2 = 10

Answer: Option C

TarunKumar1234 · Answer

Number of ways in which 5 A’ and 6 B’s can be arranged in a row which reads the same backwards and forwards, is

There are total 11 places to fill, for symmetry, we need to fill middle place first.
We have A = odd number, so one A will be in between and 2 As on left and 2 As on right side.
B = 3 left side and 3 right side

So, required combination = 5!/(2!*3!) =10 ways

I think C.

Fdambro294 · Answer

3 minutes to Brute Force this baby with a little bit of logic attached. Looking forward to looking at the more effective ways of solving this problem

(1st) because we have 11 spaces and 5 A's and 6 B's - in order for the Letters to read the Same Backwards and Forwards, we MUST make the Median Letter (6th Space) the Letter A. if we make the 6th Letter B, then the best we can hope for is to have 3 A's on one side of the Median and 2 A's on the other side of the Median. We simply will not have enough A's and B' to have to arrangement read the same forwards and backwards.

Because there is 1 Less A, we can never get an arrangement to read the same backwards and forwards if the Letter B occupies the 6th spot out of of the 11 total spots.

(2nd)whatever we choose for spot 1 -----> must be the SAME Letter for spot 11.

thus, once we have chosen a Letter for spot 1 ------> spot 11 MUST automatically contain that Letter

the same logic applies to: Spots 2 and 10 ; Spots 3 and 9 ; Spots 4 and 8 ; Spots 5 and 7

so we can start out with A in the 6th spot and we need to examine ONLY the first 5 spots to determine the no. of possibilities (b/c once we choose these spots, the last 5 spots will automatically be chosen.

__ ; ___ ; ___ ; ___ ; ___ ; A

furthermore, since we have 5 A's and 6 B's:

we must have 2 A's and 3 B's occupy these first 5 spots so that we have an equivalent number of 2 A's and 3 B's in the slots 7 through 11, inclusive, and can have an arrangement that reads the same forwards and backwards.

(3rd)Brute Force the possibilities

Start with: B -- B -- B -- A -- A ----> 1 way

then we can shift 1 A to the Left: B -- B -- A -- B -- A

now we can find all the possibilities with the FIRST 3 Letters being 'B - B- A": the A and B can switch places ---> 2! = 2 ways

then, we can shift the A to the Left Again: B -- A-- B -- B -- A

at this point, we can find all the possibilities with the FIRST 2 Letters being: B -- A

the second 3 Letters can be arranged in: 3!/2! = 3 ways

then, we can shift the A to the Left Again so that the first letter is A: A -- B -- B -- B --- A

we can find all the possibilities with the FIRST 2 Letters being: A -- B

the second 3 Letters can be arranged in: 3!/2! = 3 ways

LAST Arrangement not counted: First 2 Letters being: A - A

A - A - B - B - B -----------> 1 way

Total Number of Arrangements = 1 + 2 + 3 + 3 + 1 = 10 Total Ways

Answer: 10

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Number of ways in which 5 A’ and 6 B’s can be arranged in a row which

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