किसी घन के प्रत्येक फलक को काले या सफेद रंग से रंगा जा सकता है । उस घन को कितने विभिन्न तरीकों से रंगा जा सकता है?
सही उत्तर (B) 10 है।
व्याख्या:
A cube has 6 faces. Each face can be painted in 2 colors (black or white). If we consider unlabelled faces of a cube, we use Burnside's Lemma or direct combinatorial counting taking into account rotational symmetry of a cube (which has 24 rotations). However, let's recall the standard formula for coloring faces of a cube with k colors under rotational symmetry. For 2 colors and 6 faces, the number of distinct ways by Burnside's Lemma is given by (1/24) * sum(k to the power of cycles in each rotation). The rotations of a cube fall into 5 conjugacy classes: 1 identity rotation (6 cycles, 2 to the power of 6 = 64), 6 quarter-turn face rotations (3 cycles, 2 to the power of 3 = 8 -> 6 x 8 = 48), 3 half-turn face rotations (4 cycles, 2 to the power of 4 = 16 -> 3 x 16 = 48), 8 diagonal face-corner or 120-degree vertex rotations (2 cycles, 2 to the power of 2 = 4 -> 8 x 4 = 32), 6 edge-to-edge 180-degree rotations (3 cycles, 2 to the power of 3 = 8 -> 6 x 8 = 48). Total sum = 64 + 48 + 48 + 32 + 48 = 240. Dividing by the number of rotational symmetries (24): 240 / 24 = 10 ways? Wait, let's re-add: 64 + 48 = 112; 112 + 48 = 160; 160 + 32 = 192; 192 + 48 = 240. 240 / 24 = 10. Wait, is the answer 10 or 11? Let's check standard textbook results for coloring faces of a cube with 2 colors: 10 distinct patterns. Let's check options: 9, 10, 11, 12. Option 10 corresponds to index 1.
In English (Question & Model Answer)
Each face of a cube can be painted in black or white colours. In how many different ways can the cube be painted?
The correct answer is (B) 10.
Explanation:
A cube has 6 faces. Each face can be painted in 2 colors (black or white). If we consider unlabelled faces of a cube, we use Burnside's Lemma or direct combinatorial counting taking into account rotational symmetry of a cube (which has 24 rotations). However, let's recall the standard formula for coloring faces of a cube with k colors under rotational symmetry. For 2 colors and 6 faces, the number of distinct ways by Burnside's Lemma is given by (1/24) * sum(k to the power of cycles in each rotation). The rotations of a cube fall into 5 conjugacy classes: 1 identity rotation (6 cycles, 2 to the power of 6 = 64), 6 quarter-turn face rotations (3 cycles, 2 to the power of 3 = 8 -> 6 x 8 = 48), 3 half-turn face rotations (4 cycles, 2 to the power of 4 = 16 -> 3 x 16 = 48), 8 diagonal face-corner or 120-degree vertex rotations (2 cycles, 2 to the power of 2 = 4 -> 8 x 4 = 32), 6 edge-to-edge 180-degree rotations (3 cycles, 2 to the power of 3 = 8 -> 6 x 8 = 48). Total sum = 64 + 48 + 48 + 32 + 48 = 240. Dividing by the number of rotational symmetries (24): 240 / 24 = 10 ways? Wait, let's re-add: 64 + 48 = 112; 112 + 48 = 160; 160 + 32 = 192; 192 + 48 = 240. 240 / 24 = 10. Wait, is the answer 10 or 11? Let's check standard textbook results for coloring faces of a cube with 2 colors: 10 distinct patterns. Let's check options: 9, 10, 11, 12. Option 10 corresponds to index 1.