+0  
 
+1
83
6
avatar

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
Sort: 

5+0 Answers

 #1
avatar+89804 
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
avatar
+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.laugh

Guest Jul 15, 2017
 #3
avatar+89804 
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
avatar
+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
avatar+89804 
+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

8 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details