In how many ways can the numbers be arranged on a dice such that 1 and 6, 2…

Concept:

When numbers must be arranged on the faces of a die (an object with three pairs of opposite faces) subject to a fixed pairing of opposite faces, the count splits into two independent choices: (1) which pair of numbers is assigned to which axis of the die, and (2) for each axis, which of the two numbers in that pair faces which direction. Because these choices are made independently for each axis, the total count is (ways to assign pairs to axes) multiplied by (orientation choices per axis, multiplied across all axes). Here, an arrangement means a specific assignment of numbers to the six named faces (top, bottom, front, back, left, right); two assignments related by physically rotating the die are counted as different arrangements.

Application:

1. The dice has three pairs of opposite faces (three axes): top-bottom, front-back, left-right.

2. The pairs (1,6), (2,5), (3,4) must be assigned to these three axes. This assignment can be done in 3! = 6 ways.

3. For each axis, once a pair is assigned, the two numbers can be placed in 2 ways (either number on either of the two opposite faces of that axis). Since there are 3 axes, this gives 2 × 2 × 2 = 23 = 8 ways.

4. By the multiplication principle, the total number of arrangements = 6 × 8 = 48.

Cross-check:

Build up the same count axis by axis in a different order: choose the pair for the first axis (3 choices) and its orientation (2 choices) giving 6; choose the pair for the second axis from the remaining 2 pairs (2 choices) and its orientation (2 choices) giving 4; the last pair automatically goes to the third axis, with 2 orientation choices. Multiplying 6 × 4 × 2 = 48, confirming the result.

Hence, the correct answer is 48.