How many different ways of planting the flowers

combinationscombinatoricsdiscrete mathematics

Peter's neighbour, Paul likes tulips. He would like to plant 2 white, 5 red and 6 black tulips in a row in a way such that a red and a black tulip cannot be next to each other. How many different ways can he design the row?

I count two cases:

…w…w…

bwrwb in which wrw takes 7 positions and it can start from 7 different places.

rwbwr wbw takes 8 positions and can start from 6 different places.
…ww…

red ww black

black ww red
Which is 2 ways

So in Total I got $7+6+2=15$ ways.
But the answer is 33 ways.
I can’t think of any other ways of arrangement…

Best Answer

As there are only two white tulips that can act as separators between black and red tulips, there are two possibilities -

i) First place all black tulips together and all red tulips together. There are only two ways to do so -

BBBBBBRRRRR OR RRRRRBBBBBB

Now we must place one white tulip between them as separator. The other white tulip can be in any of the $12$ places between them or at either end.

That leads to $2 \cdot 12 = 24$ ways.

ii) We have all black tulips together but reds are on both sides or vice versa. In this case we need two separators and both white tulip positions get fixed.

When we have BBBBBB together, any number of red tulips between $1 - 4$ can be on the left and the remaining to the right. When we have RRRRR together, any number of black tulips between $1 - 5$ can be on the left and the remaining to the right.

That adds to $4 + 5 = 9$ ways.

Total number of ways $ = 24 + 9 = 33$

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Your instructor is likely assessing your ability to recognize that this is equivalent to problems like counting the number of distinct words that can be written using all of the letters in 'MISSISSIPPI'.

Apply identical reasoning to this situation: $$\frac{10!}{3!2!2!1!1!1!}$$

Matching squares

Your answers to A, B, C are correct. Part D is substantially more complicated. Here is one way to do it.

Lets name the squares $0, 1, 2, \dots, 9$ and also color them red or green in a checkerboard pattern, as follows:

   R0
R1 G2 R3
G4 R5 G6
R7 G8 R9

We will count the ways to place the four black tokens s.t. no two of them touch on a side. The red/green coloring helps the counting.

All $4$ in red squares: There are ${6 \choose 4} = 15$ ways.
All $4$ in green squares: There is only $1$ way.
$3$ red, $1$ green: If G4, G6 or G8 is picked, that rules out $3$ red squares and the black tokens must go in the other $3$ red squares. If G2 is picked, that rules out $4$ red squares and it becomes impossible to occupy $3$ red squares. Total no. of ways $= 3$.
$3$ green, $1$ red: If G4, G6 or G8 is unused, the used green squares would eliminate all red squares. If G2 is unused, the only remaining red square is R0. So there is only $1$ way.
$2$ green, $2$ red: Further case analysis on which two green squares are used:
- G2G4 or G2G6 or G4G6 would eliminate $5$ red squares;
- G2G8 would eliminate all $6$ red squares;
- G4G8 or G6G8 would eliminate $4$ red squares, so in each case the $2$ red squares are forced;
- Total no. of ways $=2$ (G4G8 or G6G8).

Therefore in total there are $15 + 1 + 3 + 1 + 2 = 22$ ways to place the $4$ black tokens. Once the black tokens are placed, we just need to pick which $3$ of the remaining $6$ squares will have white tokens. So the final answer to part D is:

$$22 \times {6 \choose 3} = 22 \times 20 = 440 $$

Best Answer

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