SOLUTION: In a row of 6 counters, 3 are red, 2 are blue and 1 is white. Find the number of arrangements in which no red counter is next to another red counter. Let the counter be num

Algebra ->  Test -> SOLUTION: In a row of 6 counters, 3 are red, 2 are blue and 1 is white. Find the number of arrangements in which no red counter is next to another red counter. Let the counter be num      Log On


   

Question 1063942: In a row of 6 counters, 3 are red, 2 are blue and 1 is white.
Find the number of arrangements in which no red counter is
next to another red counter.

Let the counter be numbered is 1,2,3,4,5,6.So the only places for red ones are
(1,3,5) (2,4,6) (1,4,6)
I started this way how to continue?
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
In combinatorics problems like this one, it is often difficult to enumerate all combinations (this one is possible because the numbers are small, but for problems involving huge numbers of combinations, you will need to reason it out, as enumeration will often be impractical). I will use R,B,W to indicate red, blue, and white, respectively. Below I show you how to solve it if each counter of a given color is identical to others of the same color, and then how to solve even if counters of one color are unique.
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Assuming the counters of the same color are identical:
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To keep the red counters apart by at least one other counter, notice that 2 blue and 1 white provide the separation:    B,B,W   

Now, for this one arrangement of B,B,W, you have 4 spaces to choose for the 3 red counters:
      1 B 2 B 3 W 4
Hence, you can fit the red ones in in 4C3 = 4!/(3!1!) = 4 ways.  (you are choosing 3 spaces out of 4 to place R)

But the BBW can be rearranged: the W can be placed amongst  the B's in 3C1 = 3 unique ways (the way shown and two additional) so that means we have a total of  4x3 = +highlight%2812%29 ways.

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If each counter is unique:
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If all 6 of the counters are unique, then we could label the red counters R1,R2,R3  and you'd have many more combinations because R1,R2,R3 is different than, say, R2,R1,R3.

For the selection of spaces between B1,B2,W — instead of 4C3 you'd have  (4C3)*(3!) because you are picking 3 of 4 spaces as before, but then you can arrange R1,R2,R3  in  3! ways for each one of those selections.

Of course, you similarly  have more arrangements of  B1,B2,W  than for  B,B,W.  The uniqueness of B1,B2 creates 2! arrangements for each selection of where you place W:
                  2!*(3C1) = 6.             

So putting it all together (4C3)*(3!)*(2!)*(3C1) = 4*6*2*3 = +highlight%28144%29+ ways, if each counter is unique.