1. ## Pigeonhole problem

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

Prove: Let $n_1, n_2, . . . , n_t \in N^+$
. If $n_1 + n_2 + . . . + n_t-t + 1$ Objects are laid in t
Pigeonholes then there's at least one $i \in \{1, . . ., t\}$
so that the i-th pigeonhole has at least $n_i$ objects
in it

3. ## Re: Pigeonhole problem

the idea is very simple. to "keep track", let's call the number of objects distributed in the i-th pigeonhole Mi.

first we check M1, by counting the objects in the first pigeonhole. if M1 ≥ n1, we are done, it is the pigeonhole the theorem claims exists.

but maybe not. so we move on to M2. if M2 ≥ n2, again, we are good, we can use M2 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 ni.

how many objects do we have in all?

n1 + n2 +...+ nt - t + 1.

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

(n1 - 1) + (n2 - 1) +...+ (nt-1 - 1) = n1 + n2 +...+ nt-1 - (t-1).

this means that there is at LEAST:

[n1 + n2 +...+ nt - t + 1] - [n1 + n2 +...+ nt-1 - (t-1)] = nt objects in the last pigeonhole.

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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 n1 + n2 - 1 objects in 2 pigeonholes.

if the first pigeonhole has n1 or more objects, we're done. if not, it has at most n1 - 1. objects. that means there's at least n2 objects left over which have to be in the 2nd pigeonhole.

the same logic applies to 3 pigeonholes and n1 + n2 + n3 - 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 n1+..+nt - t + 1 = n1+..+nt - (t-1) objects?

if we had subtracted from the sum of the ni, t objects, it would be possible to put ni-1 objects in each pigeonhole, summing to a total of n1+...+nt - t objects.

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