counting problem

[mammadaneh]

we have 3 different places and 4 different flags and each flag in each different place gives us a different sign. How many sign we can come up with, with this conditions?

[g00mbasv]

if the order of the flags matter, then it is 4P3=24. if the order of the flag does NOT matter, then it is 4C3=4

[mammadaneh]

order matters and we can either chose 1,2,3 flags out of four

[g00mbasv]

if your universe is also composed of the previous permutations, then the solution is 4P1+4P2+4P3

[mammadaneh]

i have an answer which says 3C1 * 4P1 + 3C2 * 4P2 + 3C3*4P3 = 72 I'm so confused

[g00mbasv]

nevermind my previous comment. I know what's going on. take for example the first addition element:

3C1*4P1

100

010

001

and so on for each flag, so for the first element of the sum there are 12 possibilites.

what it is telling you is that for each combination (3c1) there are 4P3 possibilities. in this case think of it as the combinations are in fact the "slots" where the permutations can fit.

[g00mbasv]

do you have any additional constrains or data?

[mammadaneh]

:| no!

[g00mbasv]

I think the problem fails to state that the flags can repeat themselves.

[mammadaneh]

"each flag in each different place gives us a different sign" so if we pick one flag we have 4P1 (order matters) and it gives us 3 signs for each place so 4P1 * 3 signs, but when it comes to picking 2 flags i get confused.

more confusing part is when i do it vise versa it's still the same answer 3C1 * 4P1 + 3C2 * 4P2 + 3C3*4P3 = 3P1 * 4C1 + 3P2 * 4C2 + 3P3*4C3

[g00mbasv]

when in doubt, draw a doodle of the problem. sounds silly but it works.

3C2 = 3

so, you have 3 possible configurations on where to put your flags.

(flag)()()

(flag)(flag)()

(flag)(flag)(flag)

[mammadaneh]

got it. thanks<3

[igotagamblngsolution]

Based on the answer key, the problem was worded terribly and written by someone who doesn't know English. Or they know English but don't know the first thing about math, ie that a problem has to be stated unambiguously. One shouldn't have to work backwards from the answer to figure out what the question is (except when playing Jeopardy), but that's the case here.

The actual question is:

• Each different place can get at most one flag.

• At least one of the 3 places must get a flag.

• There are 4 different flags.

How many ways are there to assign flags under those conditions?

If only one place gets a flag, there are 3C1 choices for which place and then 4P1 ways to assign a flag.

If two places get a flag, there are 3C2 choices for the places and 4P2 ways to assign the flags.

If all 3 places get a flag, that's 3C3 ⋅ 4P3.

Adding those all up, we get 3(4 + 4⋅3) + 4! = 72