How many people do you need to guarantee that two of them have the same initals? 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?
$26 \cdot 26$ 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.
The first two intials can be &quot;$aa,ab,ac,ad,\ldots,zw,zx,zy,zz$&quot;.
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 $676+1 = 677$.