Question

A good thinking problem.
Find the number of ways in putting n different objects in m different boxes. (any number of objects in any box)

Answer 1

are objects and boxes identical??

Answer 2

Nope.
Distinct objects. And all boxes different too.

Answer 3

\( m^n \)

Answer 4

@FoolForMath I thought the same. But apparently in the answer ( which is a series) some cases have been deducted.

Answer 5

@siddhantsharan: what do you mean?

Answer 6

As in the anser is m^n - (something)

Answer 7

*Answer

Answer 8

Okay then your question is not stated correctly.

Answer 9

you are subtracting that means no boxes could be left empty.

Answer 10

Are you sure as to subtracting being incorrect.
Also there was a hint : [ Think w.r.t. inculsion exclusion principle]
Annnd how did you get the second series?

Answer 11

The series you are looking for is \[ \sum_{j=0}^m (-1)^{j}{m \choose j} (m-j)^n \]

Answer 12

I am very sure. This is elementary mutual inclusion-exclusion. Expand that series you will get what you are looking for.

Answer 13

Okay. Thanks a lot.
But why that serries. How is it derieved?

Answer 14

How do you derive such series?

Answer 15

Annnd this subtraction (of that series) is to my question or is it when no box is empty?

Answer 16

Hiatus. Lunch time :)

Answer 17

im trying to understand the question. Is this the same as asking how many solutions does the equation:\[x_1+x_2+x_3+\cdots+x_m=n\]have, where each x_i must be non-negative? (think of each box as one of the variables, and the sum of the number of objects in each box must be n).

Answer 18

I think so.

Answer 19

Wow, joemath. I love the way you think.

Answer 20

If that is the case, then the answer is:\[\left(\begin{matrix}n+m-1 \\ m-1\end{matrix}\right)\]

Answer 21

That is not the case :)

Answer 22

How that is not the case? wait... yeah objects aren't identical.

Answer 23

That's right :)