Question is: In how many ways can 4 postcards be mailed into 3 mail boxes?
This is a topic IGCSE 0606 additional maths: Counting and the binomial expansion; The product principle)
There are 3 unique mail boxes and 4 unique postcards. (This is all from my understanding because the question is stated exactly as above; wording of word questions in a nutshell) In the first mail box, you have 5 options: Putting 0 postcard, 1 postcard, 2 postcards... 4 postcards. And then your option narrows as the number of postcard decreases.
From my understanding: Unique postcards mean 2 pair of postcards differ from another pair of postcards.
The question before this one:
In how many ways can 2 postcards be mailed into 2 mail boxes: Answer is 4
In how many ways can 2 postcards be mailed into 3 mail boxes: Answer is 9
The answer for the last one (Which is the one im asking for help) is (SPOILER): 10 + 2 + 20 + 35 + 4 + 6 + 4.
I have no idea how to work this question out. Another pattern that I notice is that (SPOILER) the answer for all 3 questions are (The mailboxes) to the power of the (Postcard).
I can provide more info is needed.
*I made bold spoiler cause some people might want to work basic high school question out :P*
Lets see
4 post cards
3 letter boxes.
All 4 in the same letter box, that is 3 ways.
3 in one and 1 in another and zero in the last
Well there is 4 ways to chose the odd postcard and then 3 boxes it can go in that is 4*3=12 then there is 2 boxes to chose from for the others so that is 2*12=12 ways
3 in one and 1 in another. 24 ways
2 in one and 2 in another and one empty.
Well there are 3 ways to choose the empty box. Then 4C2 ways to chose 2 that is 6 ways and they cand go in either of 2 boxes so that is 3*6*2 = 36 ways.
2 in one and 2 in another and one empty. = 36 ways.
1 in 2 boxes and 2 in the other box.
4C2=6 ways to chose the 2 together and they can go into one of the 3 boxes so that is 6*3=18ways
then each of the others can go in either remaining box, that is 2 ways. so altogether that is 18*2=36 ways
1 in 2 boxes and 2 in the other box =36 ways
So I get 3+24+36+36 = 99 ways
1 in 2 boxes and 2 in the other box.
There are amazingly talented persons in this world: could you explain how to put one post card into two post poxes.
huh :)
could you explain how to put one post card into two post boxes
box 1 - postcard
box2 - postcard
box3 - 2postcards
Oh I get you mr/ms Smart a**e :D
It is easy, you tear it in half and put a half into each letter box. Done :)))
I must be one of those talented people you spoke of :)
Actually I just solved it. You have 4 postcards and 3 mail boxes. We need to treat each postcards as an entity of its own. 1 Postcard has 3 options, to either go to mailbox 1, mailbox 2, or the third mailbox. That is 3 options. Since there is 4 postcards. It is then, 3 options 4 times. Translate to 3 * 3 * 3 * 3, or simply: 3^4.
I liked our guests logic so I decided to recount to work out where I went wrong.
This is what I found :)
I double counted for 2 in one and 2 in another, which meant my answer was 18 too big.
Double counting is a common mistake with these typers of questions which is why it is better to use logic like our guests if you can..
Thanks Guest - I am really pleased that you showed me your straight forward way to do it :)