Take a single pair each of Aces, 2s, 3s, and 4s from a deck of cards for a total of 8 cards. Arrange the cards in a row, such that one card is between the aces, two cards are between the 2s, three cards are between the 3s, and four cards are between the 4s.
Extension 1: Find solutions for 3 pairs (six cards: Aces, 2s and 3s) to 8 pairs (16 cards: Aces, 2s, ..., 8s).
Extension 2: Is there an upper bound? Can you find some N>8, such that N pairs cannot be arranged in this way?
Source: Reader Jimmy Chng Gim Hong, though I've seen it elsewhere.
3A2A32 4A3A2432 72632453764A5A 72632853764A5A84 I don't think there is an upper bound; you get more freedom to arrange the different pairs as you increase the number of pairs. However, there is a parity restriction on the numbers of pairs you can use. Each odd pair (like Aces or 3s) must go in two positions which are both even or both odd if you number all the positions sequentially. Each even pair goes in an odd position and an even position. Thus, if you have an odd number of odd pairs (as with 5 or 6) then you need an unbalanced number of even and odd positions (at least two more of one type) but in fact you actually have the same number of even and odd positions. Therefore it can only be done for numbers of pairs equivalent to 3 or 0 mod 4.