jax91 wrote:
RJAGmat wrote:
There are 8 red chips and 2 blue ones.
When arranged in a row, they form a certain color pattern, for example RRBRRBRRRR. How many different color patterns are possible?
We can ignore the red chips.
To place 2 blue chips in 10 slots we get -> 10P2 = 10!/(10-2)! = 10!/8! = 10x9 = 90
Further we need to divide this by 2 as the formula considers both the blue chips to be unique, which they are not.
So 45.
Hi jax91, I did this earlier but I think it is wrong.
Consider this point, You are dividing the permutations by half because of arrangements of these kinds:-
BBRRRRRRRR AND RRRRRRRRBB . These color patterns are the same.
But there will be some arrangements which will be palindromes, such as:-
BRRRRRRRRB and these will not come twice because which Blue pack is kept at the first and the last position doesn't matter. This will be true for the Blue packs coming at the following positions -> (1,10), (2,8), (3,7), (4,6) and (5,5). So, a total of 5 palindromes.
Here is how I then solved the question,
Case 1 - Palindromes :-
As said earlier, a total of 5.
Case 2 - Number of permutations in which the chips are not equidistant from each other :-
In this case you have to first consider the 10 available places made of 5 left and 5 right positions. Now, the side selected, whether left or right doesn't matter for the following example as the reason -> BB placed at (1,2) positions is same as when placed at (9,10) positions. This case has two sub-cases as follows :-
1. When both of the Blues are placed in the first or last five positions :-
5C2 = 10. (Thus, this case considers the following arrangements as the same -> BBRRRRRRRR and RRRRRRRRBB).
2. When one is placed in the left/right half and the other in the right/left half (Note: at differently distant positions) :-
(5C1 X 4C1)/2 = 10.
Divided by 2 as arrangements such as these are the same -> RBRRRRRRRB and BRRRRRRRBR .
This gives us a total of 5+10+10 = 25 color patterns.
I hope its right. Please point out any mistakes.
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