Combinatorics Word Problem

[PeterRobinson44]

There is a series of 3 docks on the bank of a river. Each dock has a flag attached to it. The flags are drawn from a store of 5 red, 4 white, and 3 blue flags and are labeled F1 through F12.

How many different ways can flags be arranged on the docks if the first flag must be red and the second flag must be white?


a. 200

b. 140

c. 240

d. 90


I understand that we are more than likely using combinatorics, C(n,r) for this particular problem. I would really like to learn the appropriate steps to answer this problem. I want to get this answer on my own, but I just want to make sure that I am doing it in the correct manner.

Is n equal to 12 because of the 12 flags?

But I am confused about r. I feel like it could be many values.

Any insight would be greatly appreciated.

 

[Dr.Peterson]

There is a series of 3 docks on the bank of a river. Each dock has a flag attached to it. The flags are drawn from a store of 5 red, 4 white, and 3 blue flags and are labeled F1 through F12.

How many different ways can flags be arranged on the docks if the first flag must be red and the second flag must be white?

a. 200
b. 140
c. 240
d. 90

I understand that we are more than likely using combinatorics, C(n,r) for this particular problem. I would really like to learn the appropriate steps to answer this problem. I want to get this answer on my own, but I just want to make sure that I am doing it in the correct manner.

Is n equal to 12 because of the 12 flags?

But I am confused about r. I feel like it could be many values.

Combinatorics doesn't mean always using C(n,r). What it always requires is careful thinking, and choosing tools from a toolbox that also includes permutations, multiplication, addition, subtraction, and some interesting tricks. Don't think you just have to pick n and r and you'll be done!

Combinations, C(n,r), would mean choosing a subset of r from a set of n, without regard to arrangement. Here, the problem is all about arrangements! The "multiplication principle" is the way to go.

I'd think of the problem as filling in three blanks, _ _ _. How many ways are there to fill in the first? Once you've done that, how many ways are there to fill in the second? And so on.

Note that if the flags weren't labeled, we'd consider them indistinguishable, and we'd be filling in the blanks with mere colors, R, W, B. Since they are labeled, they are distinguishable, and we can just call the flags 1 through 12, where 1 - 5 are red, 6 - 9 are white, and 10 - 12 are blue.

 

[PeterRobinson44]

Combinatorics doesn't mean always using C(n,r). What it always requires is careful thinking, and choosing tools from a toolbox that also includes permutations, multiplication, addition, subtraction, and some interesting tricks. Don't think you just have to pick n and r and you'll be done!

Combinations, C(n,r), would mean choosing a subset of r from a set of n, without regard to arrangement. Here, the problem is all about arrangements! The "multiplication principle" is the way to go.

I'd think of the problem as filling in three blanks, _ _ _. How many ways are there to fill in the first? Once you've done that, how many ways are there to fill in the second? And so on.

Note that if the flags weren't labeled, we'd consider them indistinguishable, and we'd be filling in the blanks with mere colors, R, W, B. Since they are labeled, they are distinguishable, and we can just call the flags 1 through 12, where 1 - 5 are red, 6 - 9 are white, and 10 - 12 are blue.

I tried to write it out. Probably wrong but I am trying to figure it out.

If I have the Red (1) then next would be White (6,7,8, or 9) and then after that, it would be (10-12) blue.
1 then 6 and then any combination 10,11,12 = 3 possibilites
1 then 7 and then any combination 10,11,12 = 3 possibilites
1 then 8 and then any combination 10,11,12 = 3 possibilites
1 then 9 and then any combination 10,11,12 = 3 possibilites
= 12 possibilites

If I have the Red (2) then next would be White (6,7,8, or 9) and then after that, it would be (10-12) blue.
2 then 6 and then any combination 10,11,12 = 3 possibilities
2 then 7 and then any combination 10,11,12 = 3 possibilities
2 then 8 and then any combination 10,11,12 = 3 possibilities
2 then 9 and then any combination 10,11,12 = 3 possibilities
= 12 possibilities

If I have the Red (3) then next would be White (6,7,8, or 9) and then after that, it would be (10-12) blue.
3 then 6 and then any combination 10,11,12 = 3 possibilities
3 then 7 and then any combination 10,11,12 = 3 possibilities
3 then 8 and then any combination 10,11,12 = 3 possibilities
3 then 9 and then any combination 10,11,12 = 3 possibilities
= 12 possibilities

If I have the Red (4) then next would be White (6,7,8, or 9) and then after that, it would be (10-12) blue.
4 then 6 and then any combination 10,11,12 = 3 possibilities
4 then 7 and then any combination 10,11,12 = 3 possibilities
4 then 8 and then any combination 10,11,12 = 3 possibilities
4 then 9 and then any combination 10,11,12 = 3 possibilities
= 12 possibilities

If I have the Red (5) then next would be White (6,7,8, or 9) and then after that, it would be (10-12) blue.
5 then 6 and then any combination 10,11,12 = 3 possibilities
5 then 7 and then any combination 10,11,12 = 3 possibilities
5 then 8 and then any combination 10,11,12 = 3 possibilities
5 then 9 and then any combination 10,11,12 = 3 possibilities
= 12 possibilities

Added all together 12 + 12 + 12 + 12 + 12 = 60 possibilities.

Am I on the right track or not close? There is probably an easier way, but I'm really trying.

 

[pka]

There is a series of 3 docks on the bank of a river. Each dock has a flag attached to it. The flags are drawn from a store of 5 red, 4 white, and 3 blue flags and are labeled F1 through F12.

How many different ways can flags be arranged on the docks if the first flag must be red and the second flag must be white?
a. 200_____b. 140_____c. 240 _____d. 90

Note that if the flags weren't labeled, we'd consider them indistinguishable, and we'd be filling in the blanks with mere colors, R, W, B. Since they are labeled, they are distinguishable, and we can just call the flags 1 through 12, where 1 - 5 are red, 6 - 9 are white, and 10 - 12 are blue.

I think that the directions are really confusing. To me:
1) it is not clear if the red flags are numbered \(F_1\,\cdots,F_5~?\) the white, \(F_6\cdots, F_9\) and blue \(F_{10},\cdots,F_{12}\)
2) Is there one flag per dock? The OP says "Each dock has a flag attached to it." if that is the case \(5\cdot 4\cdot 3=60\) which is not an option.
3) On the hand the phrase " first flag must be red and the second flag must be white " seems to me to suggest that each dock has three flags, one of each colour. That suggests \([5\cdot 4\cdot 3=60] ~+~ [4\cdot 3\cdot 2=24]~+~[3\cdot 2\cdot 1=6]\).

 

[PeterRobinson44]

I think that the directions are really confusing. To me:
1) it is not clear if the red flags are numbered \(F_1\,\cdots,F_5~?\) the white, \(F_6\cdots, F_9\) and blue \(F_{10},\cdots,F_{12}\)
2) Is there one flag per dock? The OP says "Each dock has a flag attached to it." if that is the case \(5\cdot 4\cdot 3=60\) which is not an option.
3) On the hand the phrase " first flag must be red and the second flag must be white " seems to me to suggest that each dock has three flags, one of each colour. That suggests \([5\cdot 4\cdot 3=60] ~+~ [4\cdot 3\cdot 2=24]~+~[3\cdot 2\cdot 1=6]\).

Yes, it is, unfortunately, that's exactly what the question says. It definitely could have been worded better. But your way definitely makes sense if that is what the question is asking for because of it being worded differently. I'm gonna look more into this problem.

 

[Dr.Peterson]

I get one of the answers listed when I take it as three flags in a row (one per dock), where the first is one of the 5 red ones, the second is one of the 4 white ones, and the third is one of the 10 flags remaining after that choice.

It doesn't really matter which numbers go with which color, so you can take the in the order given if you want.

 

[Dr.Peterson]

I tried to write it out. Probably wrong but I am trying to figure it out.

If I have the Red (1) then next would be White (6,7,8, or 9) and then after that, it would be (10-12) blue.
1 then 6 and then any combination 10,11,12 = 3 possibilites
1 then 7 and then any combination 10,11,12 = 3 possibilites
1 then 8 and then any combination 10,11,12 = 3 possibilites
1 then 9 and then any combination 10,11,12 = 3 possibilites
= 12 possibilites

If I have the Red (2) then next would be White (6,7,8, or 9) and then after that, it would be (10-12) blue.
2 then 6 and then any combination 10,11,12 = 3 possibilities
2 then 7 and then any combination 10,11,12 = 3 possibilities
2 then 8 and then any combination 10,11,12 = 3 possibilities
2 then 9 and then any combination 10,11,12 = 3 possibilities
= 12 possibilities

If I have the Red (3) then next would be White (6,7,8, or 9) and then after that, it would be (10-12) blue.
3 then 6 and then any combination 10,11,12 = 3 possibilities
3 then 7 and then any combination 10,11,12 = 3 possibilities
3 then 8 and then any combination 10,11,12 = 3 possibilities
3 then 9 and then any combination 10,11,12 = 3 possibilities
= 12 possibilities

If I have the Red (4) then next would be White (6,7,8, or 9) and then after that, it would be (10-12) blue.
4 then 6 and then any combination 10,11,12 = 3 possibilities
4 then 7 and then any combination 10,11,12 = 3 possibilities
4 then 8 and then any combination 10,11,12 = 3 possibilities
4 then 9 and then any combination 10,11,12 = 3 possibilities
= 12 possibilities

If I have the Red (5) then next would be White (6,7,8, or 9) and then after that, it would be (10-12) blue.
5 then 6 and then any combination 10,11,12 = 3 possibilities
5 then 7 and then any combination 10,11,12 = 3 possibilities
5 then 8 and then any combination 10,11,12 = 3 possibilities
5 then 9 and then any combination 10,11,12 = 3 possibilities
= 12 possibilities

Added all together 12 + 12 + 12 + 12 + 12 = 60 possibilities.

Am I on the right track or not close? There is probably an easier way, but I'm really trying.

It's a lot easier; you don't need all those repetitions. (This is what multiplication is for.)

How many ways are there to pick the first flag? You've listed 5.

For each of those, how many ways are there to pick the second? In each of your groupings, you've listed 4. So how many ways are there to pick just the first two flags?

Then, what you're missing is that the third flag does not have to be blue! It can be any of the flags that haven't already been used.

 

[PeterRobinson44]

If I have the Red (1) then next would be White (6,7,8, or 9) and then after that, it would be any of the rest available.
Red(1) = 40

If I have the Red (2) then next would be White (6,7,8, or 9) and then after that, it would be any of the rest available.
Red(2) = 40

If I have the Red (3) then next would be White (6,7,8, or 9) and then after that, it would be any of the rest available.
Red(3) = 40

If I have the Red (4) then next would be White (6,7,8, or 9) and then after that, it would be any of the rest available.
Red(4) = 40

If I have the Red (5) then next would be White (6,7,8, or 9) and then after that, it would be any of the rest available.
Red(5) = 40

40 x 5 = 200. Is that correct, is 90 wrong, and is there an easier way to get the correct answer using combinatorics?

 

[Cubist]

40 x 5 = 200. Is that correct

Yes - well done!

is there an easier way to get the correct answer using combinatorics?

Please re-read this post...

It's a lot easier; you don't need all those repetitions. (This is what multiplication is for.)

Like @Dr.Peterson said, you don't need to list it all out - just multiply - ie skip to the last line that you did "40*5".

You can think of this as:-

(number of options on dock 1) * (number of options on dock 2) * (number of options on dock 3)
= (number of possible red flags) * (number of possible white flags) * (number of remaining flags)
= 5 * 4 * 10

You can do this because the "number of choices" on docks 2 and 3 don't change depending on the specific choice made on dock 1.

 

[Dr.Peterson]

If I have the Red (1) then next would be White (6,7,8, or 9) and then after that, it would be any of the rest available.
Red(1) = 40

If I have the Red (2) then next would be White (6,7,8, or 9) and then after that, it would be any of the rest available.
Red(2) = 40

If I have the Red (3) then next would be White (6,7,8, or 9) and then after that, it would be any of the rest available.
Red(3) = 40

If I have the Red (4) then next would be White (6,7,8, or 9) and then after that, it would be any of the rest available.
Red(4) = 40

If I have the Red (5) then next would be White (6,7,8, or 9) and then after that, it would be any of the rest available.
Red(5) = 40

40 x 5 = 200. Is that correct, is 90 wrong, and is there an easier way to get the correct answer using combinatorics?

The answer is 200, but you don't have to write so much, as I said.

There are 5 reds and 4 whites, so there are 20 ways to choose the first two. That leave 10 others for the third flag, and 20*10 = 200. That's all there is to it: the multiplication principle, which is the foundation of combinatorics.

You wrongly got 60 (5*4*3) before by assuming the third had to be blue. I don't see where you got 90, but that would be wrong, too.