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*

Guest Jul 15, 2017

#1**0 **

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**

Melody
Jul 15, 2017

#2**+1 **

*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.

Guest Jul 15, 2017

#3**0 **

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 :)

Melody
Jul 15, 2017

#4**+2 **

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.

Guest Jul 16, 2017

#6**+2 **

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 :)**

Melody
Jul 16, 2017