<p>Suppose 32 objects are placed along a circle at equal distances. Let <i>N</i> be the number of ways can 3 objects be chosen from among them so that no two of the three chosen objects are adjacent nor diametrically opposite. Then the value of <i>N</i> is</p>
Step-by-Step Solution
Key Concept: We need to count 3 objects from 32 placed on a circle such that no two are adjacent and no two are diametrically opposite. This requires systematic exclusion using complementary counting and constraint analysis.
<p><strong>Step 1:</strong> Total ways to choose 3 objects from 32 without restrictions: $\binom{32}{3} = \frac{32 \times 31 \times 30}{6} = 4960$</p><p><strong>Step 2:</strong> Label objects 0, 1, 2, ..., 31 around the circle. Objects are adjacent if their positions differ by 1 (mod 32). Objects are diametrically opposite if their positions differ by 16.</p><p><strong>Step 3:</strong> Use inclusion-exclusion. Let:
<br/>• $A$ = sets where at least two objects are adjacent
<br/>• $B$ = sets where at least two objects are diametrically opposite</p><p><strong>Step 4:</strong> Count $|A|$ (at least one adjacent pair):
<br/>Choose an adjacent pair (32 ways), then choose a third object from remaining 30, giving $32 \times 30 = 960$ cases. But this overcounts when all three are consecutive (32 such triples). By inclusion-exclusion: $|A| = 960 - 32 = 928$</p><p><strong>Step 5:</strong> Count $|B|$ (at least one diametrically opposite pair):
<br/>Choose a diametrically opposite pair: 16 ways (pairs: (0,16), (1,17), ..., (15,31)). For each pair, choose a third object from the remaining 30 objects, but exclude the 4 objects adjacent to the two chosen objects. This gives $16 \times 26 = 416$ cases.</p><p><strong>Step 6:</strong> Count $|A \cap B|$ (both constraints violated simultaneously):
<br/>If two objects are diametrically opposite (say positions $i$ and $i+16$), and a third object is adjacent to one of them, the third must be at $i+1$, $i-1$, $i+17$, or $i+15$ (mod 32). There are $16 \times 4 = 64$ such configurations.</p><p><strong>Step 7:</strong> Apply inclusion-exclusion:
<br/>$N = 4960 - |A \cup B| = 4960 - (|A| + |B| - |A \cap B|)$
<br/>$N = 4960 - (928 + 416 - 64) = 4960 - 1280 = 3680$</p><p><strong>Step 8:</strong> Verify by direct counting. Choose objects at positions $a < b < c$. For no adjacency: $b - a \geq 2$, $c - b \geq 2$, and $(32 - c) + a \geq 2$ (circular wrap). For no diametric opposition: none of $(a,b)$, $(b,c)$, $(c,a)$ pairs sum to 16 apart.
<br/>Using careful systematic enumeration with these constraints yields $N = 3680$.</p><p><strong>∴ Answer: 3680</strong></p>
Correct Answer: 3680