Question At a certain restaurant, Barbara will order the same number of meat and fish dishes for her family dinner. How many different platters, consisting only of meat and fish dishes, could Barbara order?
Statement 1 – The restaurant offers 5 different meat dishes.
Statement 2 – the maximum number of meat and fish dishes that Barbara can order is 8.
Solution:Qn: how many different platters are possible; Given, same no for fish and meat.
Thinking about the above: we could have Meats (M1, M2, M3, M4 ,M5, M6, M7 and so on) and Fishes (F1, F2, F3, F4, F5, F6 and so on) to select from.
S1: 5 meat options are available, the cases possible will depend on lesser of the 2 quantities. So S1 alone not enough
S2: Max total is 8, so possible combos can be: 1 meat + 1 fish dish or 2 meat + 2 fish dishes or 3 meat + 3 fish dishes or 4 meat + 4 fish dishes, but we dont know how many of each are available i.e. again size depends on item with limited availability . So S2 alone not enough
S1+S2: now we know that we have 5 Meat choices but max 8 means we have 4 Fish options to select from. At this stage we can mark C and move on.
But if you are interested in solving further (especially if you are hungry to learn more

, see that I did there):
Total cases = sum of possible cases of selecting (1 Each + 2 Each + 3 Each + 4 Each) from among 5 Meat options and 4 Fish options.
1 Each: 5c1*4c1 = 5*4 = 20
2 Each: 5c2*4c2 = 10 *6 = 60
3 Each: 5c3*4c3 = 10*4 = 40
4 Each: 5c4*4c4 = 5*1 = 5
Total possible platters is 125.
Hope this helps