In how many ways $5$ $A$'s and $6$ $B$'s be arranged in a queue so that it reads the same forward and backward?
I got that middle letter will be $A$ so that $A$ and $B$ can be arranged symmetrically. But I am struggling to get the answer.
Asked
Active
Viewed 48 times
3 Answers
0
Clearly, A has to be at the center position. Now, we need to have $2$ A's and $3$ B's on one side and other letters should be mirror image of these about the center A.
So, the required number of ways: $$\binom52 =10$$
Examples include:- BABABABABAB, BBBAAAAABBB.
-
Thanks.now it is clear – Harsh Vardhan Jan 02 '18 at 14:29
-
@HarshVardhan You can accept my answer by ticking the green mark next to my answer. – Jan 02 '18 at 14:31
0
So firstly, we have $11$ letters, $4 A's$ and $6 B's$ up for picking. The center slot is locked, so we essentially need to determine $$\frac{(11-1)}{2} = 5$$ slots, since it is symmetrical. We will hence need $2 A's$ and $3 B's$ to be on each side. Now all we need to do is to permute the number of ways the $2 A's$ and $3 B's$ can be arranged. We use $^5C_2$, since we need to choose $2$ seats from the $5$ possible seats, which is $10$. Hence, there are $10 $ways.
Rohan Shinde
- 9,737
QuIcKmAtHs
- 1,451
0
As you deduced, the middle letter must be A.
The remaining letters must be symmetric about the central A. This means 2 As and 3Bs on both sides. There are 5 letters on a given side of the central A, and not all are different.
If the 5 letters were different, there would be $5!$ permutations. To address the question posed in your title, the 2 identical As can be arranged in $2!$ ways and the 3Bs are identical and they can be arranged in $3!$ ways so the total number of permutations of the 5, given some objects are alike is $$ \frac{5!}{2!\,3!}=10 $$
In general, if you have $n$ objects, and $n_1$ are alike, and $n_2$ are alike and $n_3$ are alike $\ldots$ then the number of permutations is
$$ \frac{n!}{n_1!\,n_2!\,n_3!\ldots} $$
PM.
- 5,249