Permutations & Combinations
Number formation
Grade 11
Question:
<p>The total number of six-digit natural numbers that can be made with the digits 1, 2, 3, 4, if all digits are to appear in the same number at least once is</p>
<p>1560</p>
<p>840</p>
<p>1080</p>
<p>480</p>
Step-by-Step Solution
Key Concept: Use inclusion-exclusion principle on surjective functions: Total arrangements where each of 4 digits appears at least once in 6 positions equals arrangements minus those missing at least one digit.
<p><strong>Step 1:</strong> We have 6 positions and 4 distinct digits {1,2,3,4}. Each digit must appear at least once.</p><p><strong>Step 2:</strong> Total arrangements without restriction = 4^6 = 4096</p><p><strong>Step 3:</strong> Apply inclusion-exclusion. Let A<sub>i</sub> = arrangements missing digit i.</p><p><strong>Step 4:</strong> |A<sub>i</sub>| = 3^6 = 729 (using only 3 digits)</p><p><strong>Step 5:</strong> |A<sub>i</sub> ∩ A<sub>j</sub>| = 2^6 = 64 (using only 2 digits)</p><p><strong>Step 6:</strong> |A<sub>i</sub> ∩ A<sub>j</sub> ∩ A<sub>k</sub>| = 1^6 = 1 (using only 1 digit)</p><p><strong>Step 7:</strong> By inclusion-exclusion:</p><p>Answer = 4^6 - C(4,1)·3^6 + C(4,2)·2^6 - C(4,3)·1^6</p><p>= 4096 - 4(729) + 6(64) - 4(1)</p><p>= 4096 - 2916 + 384 - 4</p><p>= <strong>1560</strong></p><p>∴ Answer: C</p>
Correct Answer: C