Each person is allowed to buy no more than 3 tickets. So the maximum number of tickets that can be bought by the first pair in line would be 6. We are told that only 5 tickets are bought by the first people in line, so this requires one person in a pair to buy 3 and the other to buy 2. Let’s call the number of tickets available for purchase N. Since it takes a pair to buy 5 tickets, then the number of pairs=N/5.
The number of people in N/5 pairs is 2N/5. The number of people without tickets is 25-2N/5.
We need to find N. We know that 25-2N/5>0, that is, some people get no tickets. Therefore 2N/5<25, and N<125/2, that is, N<62 tickets. However, the first people in line are buying 5 tickets until they run out, so N must be divisible by 5. The nearest multiple of 5 less than 62 is 60, which means that 12 pairs (24 people) bought tickets and 1 person was left without a ticket. If N=55, then 25-2N/5=3 people without tickets; if N=50, then 25-2N/5=5 people without tickets, and so on up to N=5 (23 people get no tickets). If we assume that the maximum number of tickets is 60, then only one person in the line goes without a ticket.
The only other way to interpret this question is to assume that the first people in line bought 5 tickets each, which breaks the rule about the maximum of 3, and is therefore not an acceptable assumption. The ticket sellers do not know in advance how many tickets to produce because they don’t know the demand—unless, after seeing 25 people in line they decide on a maximum of 3 per person, making 75 tickets available and restricting sales to a maximum of 3 tickets per person. But 75 exceeds the maximum for N (N<62) and everyone would get a ticket, which is not what the question implies.