# Thread: how many for at least two to have same first, last initials

1. ## how many for at least two to have same first, last initials

How many people would have to be in a school before at least two people had to have the same first and last initials?

2. I believe this is a trick question. The probabability of two people having the same first and last initial would be 1/26^4, but that does not mean that you NEED 456976 people in the school. You could concievably have two people in the school and they both have the same initials, it just isn't as probable. The probability of two random people in the school having the same initials would be 1/456976, but the denominator isn't how many people are needed before two have the same, it is just the total possible combinations.

3. Think of all the ordered pairs of the alphabet, $26 \times 26 = 676$.
So we can have 676 people with different first and last initials.
Thus if we have 677 people we must have at least two with the same first and last initials. This is a classic pigeon hole problem.

4. Yes, the pigeonhole principle is very useful to know.

A basic problem is:

Given a group of n women and their husbands, how many people must be chosen from this group of 2n people to guarantee the set contains a married couple?.

Note how the solution to your problem is $n^{2}+1$?.

5. Originally Posted by galactus
Given a group of n women and their husbands, how many people must be chosen from this group of 2n people to guarantee the set contains a married couple?.
That's a classic? It's worded a bit oddly, isn't it? I'm pretty clear on "this group", but I'm not quite sure what "the set" means.

From a strictly wording point of view, "the set" could be the folks with which we started. In this case, we would ahve to select none to guarantee, since that is the definition ofthe group.

From a what-would-be-interesting point of view, it's clear that "the set" means the set of individuals selected from "this group".

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