Where does the $(c-1)$ come from?

Question

I received the following question:

(This is somewhat pigeonhole in reverse) A person wants to collect c identical coupons. There are n different types of coupons. How large must m be so that if this person collects m coupons, he/she is guaranteed to obtain c identical ones? (m must be a function of c and n)

The answer was revealed and it starts with:

We need the following to be true $\lceil m / n \rceil = c$

That part I understand as it relates to the pigeonhole problem.

It then goes on to list out the answer without an explanation:

$m = n(c-1) + 1$

My assumption was that it would be: $m = n \cdot c+1$

Where does the $(c-1)$ come from?

Answer 1

Suppose you start with $M = n(c-1)$, where there are $n$ different kinds of coupons. Is it possible, at this point, that you do not yet have at least $c$ coupons of any one particular type?

Yes
That is, you could have exactly $(c-1)$ coupons of each of the $n$ types, since you are assuming that $M = n(c-1)$.

However, if you now acquire one more coupon, then one of the $n$ types will then have to be pushed over the edge from having $(c-1)$ coupons of that type to having $c$ coupons of that type.

So, this explains the computation of
$[n(c-1)] + 1.$

Answer 2

This might give you a better feel for the claim: $$m=n(c-1)+1$$ Consider the extreme scenario where the person collects $c-1$ coupons of each of the $n$ types.
He has not collected $c$ coupons of any type but if he collects one more coupon, he is guaranteed to have $c$ coupons of one of the $n$ types.

Answer 3

We require the smallest $m$ such that $\lceil \frac{m}{n} \rceil = c$. This means that $\lceil \frac{m-1}{n} \rceil < c$.

Because $c$ is an integer and we know $m-1$ is maximal, we can safely remove the ceiling brackets and use an equals sign:

$$\frac{m-1}{n} = c-1$$

so that

$$m = n(c-1)+1$$