6 is the number of combinations for one evaluation item only, C(4,2) = 6; it forgets to multiply by the 3 items price, performance, and service. The question states that all combinations are compared for each of price, performance, and service, and Table 3 also provides separate rows for performance and service, so three times 6 is required.
×Incorrectイ
9 comes from 3×3, and shows up when you miscount, for example as “comparing one reference model against the remaining 3 models (3 comparisons) × 3 items.” But this question asks for the relative merits of every combination, so the comparisons B vs C, B vs D, and C vs D, which do not involve Model A, are also needed; 3 comparisons per item is not enough, and 6 are required. Hence 9 does not complete the table.
○ウ
正解One row of the table corresponds to one comparison, so the number of comparisons needed equals the number of rows the table has when it is completely filled in. First consider just one evaluation item: count the combinations of choosing 2 models out of the 4 models A to D. Order does not matter (comparing Model A with Model B and comparing Model B with Model A are one and the same comparison), so the number of combinations is C(4,2) = 4×3÷2 = 6, namely A vs B, A vs C, A vs D, B vs C, B vs D, and C vs D. The lines “Model A ＝ Model B” and “Model A ＞ Model C” listed in the price part of Table 3 show these 6 pairs in the middle of being filled in one after another. Next, the same comparison is carried out for each of the 3 evaluation items — price, performance, and service — so 6 comparisons × 3 items = 18. Therefore ウ, 18, is correct. In general, with n models and m items the number of comparisons is given by n(n−1)÷2 × m.
×Incorrectエ
32 cannot be derived by any way of counting and is far too large. Even if you distinguish order (treating comparing Model A with Model B and comparing Model B with Model A as different), you get only 4×3 = 12 pairs, which gives 36 comparisons for 3 items, not 32. One comparison is enough to settle the relative merits of the same pair of models, so counting with order distinguished is itself wrong; the correct count is 6×3 = 18.
ここがポイント
The number of combinations of choosing 2 out of n items is n(n−1)÷2. For 4 models that is 4×3÷2 = 6, and the key point is not to distinguish order, so that A vs B and B vs A are not counted twice.
When the same all-pairs comparison is repeated for several evaluation items, the total is the number of combinations per item × the number of evaluation items. In this question that is 6 × 3 = 18.
It is also worth grasping the definition given in the question — one row of the table corresponds to one task (one comparison) — since the problem is essentially about counting how many rows are needed.