Simple combinatorics,but retarded question

I'm giving a questions that is rather vague in writing here it is.
7 distinct pigeons are sent to 3 different cities A,B,C.
How many possibilities is there if for city A was sent no less than 2?

i think the answer is 2*5*5 is that correct?

>i think the answer is 2*5*5 is that correct?
No.

My bad meant to write (2*5*4)+(3*4*3)+(4*3*2)+(5*2*1)

what about (2*5*4)+(3*4*3)+(4*3*2)+(5*2*1)?

>My bad meant to write (2*5*4)+(3*4*3)+(4*3*2)+(5*2*1)
No.

[math]{9\choose 2}-{6\choose 1}[/math]

This guy is a literal retard or a troll or both

Just work out every case. How many ways when 2 are sent to A? How many when 3 are sent to A? Etc. Use summation notation :^)

is it 14?

Holy shit Veeky Forums is retarded. How many ways can you distribute them? 3^7 of course. How many of these arrangements have only 1 pigeon in town A? 7*2^6. So the answer is 3^7-7*2^6. Fucking brainlets

No it's a lot more than that.

When A gets two pigeons, there are 7choose2 ways to choose them. Then there are 5 possible numbers of pigeons for city B. For each of those numbers k, there are 5choose k ways to send them. Then the ones sent to C are determined. Add all those up. Then repeat the analysis for 3 getting sent to A, etc.

These are the type of people who solve geometry problems with integration

thats what i thought i did when i wrote (2*5*4)+(3*4*3)+(4*3*2)+(5*2*1)
for the first one if 2 is sent to A then there can be 5 pigeons that can be sent to B and then 4 to C,and so on when 3 is sent to A then...
is what im writing different from that logic?

Nah, I think this is pretty clearly a Pigeonhole Principle problem. You might be on to something though.

the pigeons are distinct. imagine each pigeon has a nametag with a different number on it. so 1,2 getting sent to A is different from 2,3 getting sent to A.

haha but what if the town of A does have no pigeons at all

>Nah
Yeah, it's correct. 3^7 total arrangements (do you agree with that part?). 7*2^6 arrangements where city A only has 1 pigeon. Subtract the two

Fuck I should kill myself. Obviously subtract that case too. So also subtract 2^7. God damn it looks like I'm the brainlet

If it was required to count the possibilities there is if for city A was sent exactly 2
then the answer would be 7*2^5 ?

No, it would be (7choose2)*2^5

That's equal to 21*2^5 right?
sorry for being a brainlet

>7choose2

How many ways are there to pick 2 objects from 7?

Well, you have 7 choices for the first one, 6 choices for the second one, and then divide by 2 possible arrangements. So it's 7*6/2=21

gratitude stranger !

Just to make sure that i understood correctly,what if it was required to count the possibilities there is if for city A was sent no less than 5
so the answer should be:
3^7 -(7choose4)*2^3 -(7choose3)*2^4 -(7choose2)*2^5 -(7choose1)*2^6 -2^7 ?

I think once you get to 5, it's easier to do it constructively (ie, count the VALID cases, instead of subtracting the invalid cases from the total). So I would do (7choose5)*2^2+(7choose6)*2+7choose7

Physics fag here

What is wrong with the second anons method?

It's correct, but you would need to calculate the cases for 2,3,4,5,6,7 pigeons at A. Which is way longer than just counting the cases with 1 and 0 pigeons and then subtracting. It's similar to the explanation in

So is this not correct

I can't even tell what the hell this is.

There will always be a nonnegative integer number of ways to send pigeons to cities so no, that is not correct.

I wrote out everything that it could be, then counted everything greater than or equal and divided by total number of answers and simplified

This isn't the right way to approach combinatorics problems. Don't write out the possibilities, you won't get anywhere with that. It takes too long and you're prone to make errors

You should subtract rather than divide.

(total solutions) - (bad solutions) = (good solutions)

So what is the correct answer

>So what is the correct answer

You have been provided with enough information to work it out on your own. Have fun, and good luck!

>So what is the correct answer
You cannot be serious. JFC

Here is another one anons
what if it was asked to count the possibilities
if no less than 5 pigeons were sent to A ,and exactly 1 pigeon was sent to b

is that (7choose6)*7 -2^6

7 choices for the pigeon that is sent to B. Then either 5 or 6 pigeons were sent to A. In the first case, there are 6 choose 5 ways, and in the second case there is only one way.

This assumes that it's possible that none of them get sent to C, which is ambiguous in this problem and in the original.

isnt that supposed to be 7*P(6,5)+7*P(6,6) ?