I'm struggling with this problem for a while now, and I just can't figure it out.

Prove: Let

. If Objects are laid in t

Pigeonholes then there's at least one

so that the i-th pigeonhole has at least objects

in it

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- Jan 15th 2013, 10:51 PMsebastianhatlPigeonhole problem
I'm struggling with this problem for a while now, and I just can't figure it out.

Prove: Let

. If Objects are laid in t

Pigeonholes then there's at least one

so that the i-th pigeonhole has at least objects

in it - Jan 16th 2013, 01:38 AMemakarovRe: Pigeonhole problem
Try a proof by contradiction.

- Jan 16th 2013, 02:20 AMDevenoRe: Pigeonhole problem
the idea is very simple. to "keep track", let's call the number of objects distributed in the i-th pigeonhole M

_{i}.

first we check M_{1}, by counting the objects in the first pigeonhole. if M_{1}≥ n_{1}, we are done, it is the pigeonhole the theorem claims exists.

but maybe not. so we move on to M_{2}. if M_{2}≥ n_{2}, again, we are good, we can use M_{2}as the desired pigeonhole.

but let's say it's just the worst possible case, the first t-1 pigeonholes all have fewer objects than the corresponding n_{i}.

how many objects do we have in all?

n_{1}+ n_{2}+...+ n_{t}- t + 1.

what's the MOST possible objects we could have come across in our first t-1 pigeonholes?

(n_{1}- 1) + (n_{2}- 1) +...+ (n_{t-1}- 1) = n_{1}+ n_{2}+...+ n_{t-1}- (t-1).

this means that there is at LEAST:

[n_{1}+ n_{2}+...+ n_{t}- t + 1] - [n_{1}+ n_{2}+...+ n_{t-1}- (t-1)] = n_{t}objects in the last pigeonhole.

*********************

it's easier to see what is happening with a small number of pigeonholes.

suppose we have only one, and we have k objects distributed in 1 pigeonhole. then that pigeonhole has at least k objects (duh!).

perhaps that's "too easy". ok, let's try TWO pigeonholes, with n_{1}+ n_{2}- 1 objects in 2 pigeonholes.

if the first pigeonhole has n_{1}or more objects, we're done. if not, it has at most n_{1}- 1. objects. that means there's at least n_{2}objects left over which have to be in the 2nd pigeonhole.

the same logic applies to 3 pigeonholes and n_{1}+ n_{2}+ n_{3}- 2 objects. rather than go through the same steps, let's pick 3 numbers, like 3,6, and 8. 3+6+8 = 17. 17 - 2 = 15.

so we have 15 objects to place in 3 pigeonholes. can we "beat the theorem" by placing less than:

3 objects in the first pigeonhole (ok, we want to get rid of "as many as we can" so we put 2. now we have 13 left).

6 objects in the 2nd pigeonhole (the most we can put is 5, right? that leaves us with...uh...8 left).

8 objects in the 3rd pigeonhole (well, shucks...we're stuck!)?

do you see why the theorem claims we have n_{1}+..+n_{t}- t + 1 = n_{1}+..+n_{t}- (t-1) objects?

if we had subtracted from the sum of the n_{i}, t objects, it would be possible to put n_{i}-1 objects in each pigeonhole, summing to a total of n_{1}+...+n_{t}- t objects.

by only subtracting t - 1 objects, we ensure at least ONE pigeonhole has n_{i}objects (the "extra one we didn't subtract").