How many people do you need to guarantee that two of them have the same initals?

Question

An auditorium has a seating capacity of 800. How many seats must be occupied to guarantee that at least two people seated in the auditorium have the same first and last initials?

I thought $26 \cdot 26$ but the answer key has $26 \cdot 26+1$ Where's the +1 coming from? Is the whole thing about the auditorium a red haring, I don't see how it matters?

Answer 1

$26 \cdot 26$ would count all possible pairs of letters. With $26 \cdot 26$ people it is possible that they all have different initials. The $+1$ ensures there exist at least two people with the same initial.

Maybe it would help to think of a smaller case. If one rolls a dice $6$ times, it's possible to roll a different number each time. However, rolling a dice $6+1=7$ times ensures that at least one number was rolled more than once.

Answer 2

It's theoretically possible for there to be $26\times26$ people all with different first/last initials...

Answer 3

It's the pigeonhole principle at work. There are $26^2$ people possible with differnt initials (From AA to ZZ). But if you add one more to the mix, then the new guy cannot have an initial not counted already. It will be one of the AA to ZZ. Hence this guarantees it.

Answer 4

The first two intials can be "$aa,ab,ac,ad,\ldots,zw,zx,zy,zz$". So, total possibilities for different types of first two intials are $26\times 26 = 676$.

And they asked us to have same first two intials for atleast $2$ children, so we can take any of the $676$ possibilities as a common first two intials. So, number of ways for at least $2$ students who have the same first two initials are $676+1 = 677$.