Lunchtime Scrambles

We know that there are $5 \times 6 = 30$ possible fruit-sandwich combinations. If we have at least $31$ students in the class, two students must have both the same fruit & the same sandwich, which is a contradiction; thus, there can be at most $30$ students.

It is also possible for there to be exactly $30$ students, since in that case we could set up a one-to-one correspondence (a bijection ) between the students and the fruit-sandwich combinations, and so avoid having more than one item in common.

Therefore, $\boxed{30}$ is the largest number of students that could be in the class.

No more than 6 students can have the same sandwich, because they each must get a different fruit. Therefore the number of students is no more than $6 \times 5 = 30$ .

On the other hand, it's possible to have 30 students subject to the condition: for each of the five sandwich choices, we give out six; and then we give a different choice of fruit and a different choice of vegetable to each of the six students.

Therefore the largest number of students that could be in the class is 30.

Let's denote choices of fruit = {a, b, c, d, e, f}, choices of sandwich = {A, B, C, D, E}, and choices of vegetables = {1, 2, 3, 4, 5, 6, 7}. We should arrange the items so that no two students have more than one item in common. Hence, here's one of the possible ways to arrange. We arrange them in a sequence (a, A, 1), (a, B, 2), ..., (a, E, 5), (b, A, 6), ..., (b, E, 3), ..., (f, E, 2). It can be observed that the pair (A, 1) does not occur twice because the LCM of 5 and 7 is 35, and there are only 6 $\times$ 5 = 30 arrangements. Same goes to other pairs in the form (sandwich, vegetables). Thus, to fulfill the given condition, the largest possible number of arrangements of the items is 30 and so is the largest number of students.

I know I'm a late solver and there's no hope for a vote at this stage. But yet I'd like to share a solution ( without solution, a problem looks odd :p ).

My solution would be easily digestive. I'm not naming a theory/principle.

Here, we can fix our gaze on sandwich first, as it has the lowest variety - chances of repeatation increases. Like: if a boy choses mutton sandwich, he can chose $1$ out of those $6$ vegetables. Another student chosing the same sandwich, cannot chose the then chosen vegetable. So, at most $6$ student can chose mutton sandwich. Same case occurs for all $5$ sandwiches.

Sum up: there's $5\times6=30$ possible food package for individuals only. That sorts out the largest number of students there.

Start off with picking the food with the least amount of possible choices, which is the sandwich with 5.

For each set of food combinations where the sandwich is the one item in common, the fruits and vegetables have to all be different from each other. Since there are 6 choices for the fruit and 7 choices for the vegetable, the maximum amount of completely unique pairs is 6. It cannot be greater than 6 because there are only 6 unique choices for fruits.

To put it simply, imagine trying to make fruit-vegetable pairs using 6 fruits and 7 vegetables. You'll end up with 6 fruit-vegetable pairs, with 1 vegetable not being used.

So the answer is 5*6=30 (5 sandwich choices and 6 completely unique fruit-vegetable pairs)

As for the fruits and vegetables both repeating when the sandwiches are different, there are 6*7=42 ways to uniquely pair up the fruits and vegetables without having both the fruits and vegetables repeat, so there's no need to change anything because of that.

The above paragraph was a lucky stroke (haha)