# Ways to put letters in postboxes

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1. Jul 31, 2015

### fireflies

1. The problem statement, all variables and given/known data

There are 7 different postbox, and 10 identical letters. How many ways can the letters put into the boxes so that there is at least one letter in a postbox?

2. Relevant equations
nCr=n!/(n-r)!r!

If M,N,O..... things can be done in m,n,o..... ways then ways of doing them together is m*n*o..

3. The attempt at a solution

First I tried to put 7 letters in 7 boxes. It can be done in 1 way only as all the letters are identical. Then I tried to put the rest three letters in the box. Let us mark the letters as A, B and C (just to describe easily). While putting A in a box, for every box there are two solutions, either yes you put or no you don't put. So ways are 2^7. But it also includes that you dont put it any boxes. So, actual ways are 2^7-1. Same for letter B and C. So total ways can be (2^7-1)^3 = 2048383 ways.

The problem solution was given a number something less than 100. I think the first step maybe a problem where I said ways to put 7 letters is 1. But then the answer is going to be bigger, not less than 100 anyways. I asked the one who gave this problem about my solution, he couldn't answer. It was on facebook and not on my id. And I don't remember how his solution was. Can anybody tell me where the problem is?

2. Jul 31, 2015

### Borg

Since you have to have one letter in each box, the number of combinations is the same as the number of ways to arrange 3 idential letters in 10 different boxes. Does that make it easier for you?

3. Jul 31, 2015

### fireflies

Yes, but how to put these three in 7 boxes (Maybe you mistook 10)? Is my second step to solve this a good approach?

4. Jul 31, 2015

### fireflies

Maybe the solution given there was

1) All three letters in same box in 7C1

2) 2 letters in a box and rest in another box in 7C1*6C1

3) 3 letters in 3 different boxes in 7C3

Then sum of case 1, 2 and 3.

This makes sense to me. But what's wrong with my concept then?

5. Jul 31, 2015

### RUber

Fireflies, in your original method, you not only are allowing for no box to get a letter from the leftover 3, but also for every box to get a letter (7>3). That is why your solution is so large.
--As you just posted--
You have the options to put:
1 additional letter into 3 boxes
2 additional letters into 1 box and 1 letter into another
3 additional letters into 1 box
This is a good way to look at the problem.

6. Jul 31, 2015

### Borg

The case that you have here is if all of the boxes are identical.

Last edited: Jul 31, 2015
7. Jul 31, 2015

### fireflies

Yes, all the boxes are not identical. That's my point.

Well, I subtracted that 1 to confirm no box is selected. But that includes all yes options too. Now I get what's the actual mistake hete

8. Jul 31, 2015

### fireflies

Then if I say in this way, can this be right-
For letter A to select a box is 7C1. Same for letter B and letter C? So, finally (7C1)^3

Or it is wrong as the letters are identical?

9. Jul 31, 2015

### Borg

Since the letters are identical, there are only 7 ways to fill a box with 3 letters. Does that make sense?

10. Jul 31, 2015

### RUber

That would be wrong, since with identical letters AC = AB, so 7^3 includes combinations with AC, AB, BC in the same box counted as distinct outcomes.

11. Jul 31, 2015

### fireflies

So, you are saying that the way with three cases (in #4) is correct? What would be the case then if the post-boxes were identical?

12. Jul 31, 2015

### fireflies

No, how? The post boxes are not identical. 3 letters in each box itself is going to be 7 ways. 2 in one box rest in other can add more choices like box no 1,2; 2,1; 1,3 etc. 3rd case will give extra choices.

13. Jul 31, 2015

### Borg

Post #5 gives you the 3 variations that you have to consider.
Post #9 gives you the answer to one of those variations.

14. Jul 31, 2015

### fireflies

Sorry, maybe I got it wrong. All three letters in a box and this can be done in 7 ways, that does make sense

15. Jul 31, 2015

### fireflies

Now I can actually get what you're trying to say :)

But I already did that in post#4 right?

16. Jul 31, 2015

### RUber

If the boxes were identical and the letters were identical you would only have:
6 boxes with 1, 1 box with 4
5 boxes with 1, 1 box with 2, 1 box with 3
4 boxes with 1, 3 boxes with 2.
That is: 3 ways to split 10 letters into 7 boxes with at least one in each box.

This is almost certainly not what the original question was looking for.

17. Jul 31, 2015

### fireflies

Since now it is more clear to me, let me try to answer post#11 by myself. Here post-boxes are identical. So, we can only do is
3 in a box in 3C3
2 in a box and rest in other in 3C2
1 in each box in 3C1,

Then add all. Am I correct?

18. Jul 31, 2015

### fireflies

I read the post later than I posted my answer. It makes me confused again. Why not we put the 7 in their case and try to combine the rest three here?

Yes, it is not an answer to the original post, but since I am having problems to work out the conditions so I think these answers are more likely to help me how to solve these actually

19. Jul 31, 2015

### Borg

You have the answer to this one. So let's look at the next one up:
Just answer this question first: How many ways can you put 2 additional letters into 1 box?

20. Jul 31, 2015

### RUber

Are you still assuming letters are identical too?

3 in a box means 3 in one box 0 in others. If boxes are identical, there is only one way to do it.
2 in one box, 1 in another, and 0 in others. This is still one case, all similar layouts are considered indistinct.
1 in each of 3 boxes and 4 empty boxes. Also, just one case.

If the letters are not identical, then you need to start from 10 letters again. I reiterate, this is not what the question was asking and would likely become complicated quickly.
( I could be wrong on these since I don't often deal with these sorts of questions)
Case 1: 4 in one box, one in each of the other 6-- 10C4 ways to choose the 4 letters that go in.
Case 2: 3 in one box, two in another, 1 in rest--10C5 ways to choose the ones going into two boxes times 5C3 ways to choose which ones go into the first.
Case 3: 2 in three boxes, 1 in the rest -- 10C6 * 6C4 * 4C2 ways

Then sum the 3 cases.