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Assuming that everyone in a particular school has three initials, find out whatis the smallest number of students in a school for which there must be at leasttwo with the same initials.

Richard
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  • Perhaps if you're having trouble with the question, don't start the title by declaring it to be 'easy' in all-caps.... –  Feb 14 '14 at 04:13
  • @Richard: How many triples of initials can you have? – user99680 Feb 14 '14 at 04:15
  • What you have you tried? – BlackAdder Feb 14 '14 at 04:34
  • I called it easy since I already had an answer in mind, but although David nailed it, I don't understand why the +1. – Richard Feb 14 '14 at 05:09
  • @Richard I didn't see your comment yesterday. Consider the simplified problem of only two names per person and there being only 2 letters per initial, say A and B. Then every person would have initials either AA, AB, BA, or BB. So four possibilities. If there were 4+1=5 people in the room, two people must have the same initials. – David P Feb 15 '14 at 12:53

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Find the number of possible unique initial combinations, and add 1. The number of possible initials, assuming capital letters and the english alphabet:

$$26\cdot 26\cdot 26=17576$$

If there were at least $17577$ students, by a concept known as the Pigeonhole Principle, (in this case, common sense really) two people must share initials.

David P
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