There is this problem in my schoolbook:
A panel is composed by 6 rectangles as shown below:
How many different ways are there of painting this panel, knowing that 2 of the rectangles must be painted white and the remaining 4 have to be painted with different colors each, which can be either green, red, yellow, blue or pink?
To solve this I would do:
- combinations of white tiles: 6*5
- combinations of different colored tiles: 5*4*3*2
- Total combinations: = (6*5)*(5*4*3*2) = 3600
However,my schoolbook says the solution is supposed to be 1800, which happens to be half of what I got.
According to the book, the solution is:
We start by choosing the 2 rectangles that are going to be painted
white and then we choose the 4 colors among the 5 with which we will
paint the remaining rectangles $^6C_2$ $*$ $^5A_4 = 1800$Another way of solving this is by choosing 4 of the 6 rectangles that
will be painted with 4 of the 5 colors, leaving the remaining ones
white. $$^6A_4*^5C_4= 1800 $$ or $$^6C_4*^5A_4= 1800$$
My questions:
- What am I doing wrong?
- What do A and C mean in this context?
- I know that the solutions mentions the formula/method used when you
don't want repetitions of the same elements regardless of the order,
but should that matter in this context? Or is the problem just not a
very good example of when to use this method?
Maybe what the author of the problem meant was to get all the possible combinations without repeating elements regardless of the order, however that is not explained properly?
There is also a chance my schoolbook is wrong, or the problem has an error, is missing information or is not explained properly. That sort of thing happens all the time . I copied the problem exactly as it is in the book.
Best Answer
"What am I doing wrong?": When you say "combinations of white tiles: $6*5$" your answer seems to be assuming that the first rectangle chosen to be white is somehow different than the second rectangle chosen to be white.
To correct this mistake, you should divide by two since you have otherwise counted every scenario twice.
"What do $A$ and $C$ mean in this context?": The binomial coefficient counting the number of ways of picking $r$ objects out of $n$ total where order doesn't matter has a number of different notations. Included in this list are $\binom{n}{r}, \frac{n!}{r!(n-r)!},~^nC_r,~_nC_r, C(n,r),\dots$
In this specific case, you have $\binom{6}{2}=~^6C_2=\frac{6!}{2!4!}=\frac{6\cdot 5}{2}=15$
The $A$ is apparently an analogous notation used for falling factorials. The number of ways of placing $r$ objects out of $n$ total where order does matter has a number of notations. Included in this list are $~_nP_r, ~^nP_r, n^{\underline{r}}, \frac{n!}{(n-r)!}$ and apparently your books uses a variation of this being $~^nA_r$.
In this specific case, you have $~^6A_4=\frac{6!}{(6-2)!}=6\cdot 5\cdot 4\cdot 3$
"...Should that matter in this context?...": It matters in so far as whatever method you use to count (there are several) you want to make sure that you count each outcome exactly once. Make sure that your method of counting doesn't allow you to reach the same outcome via two (or more) different sequence of choices.