Permutations & Combinations
Counting with Constraints
Grade 11
Question:
<p>Let <i>x</i> be the number of 6 digit numbers, the sum of whose digits is even and <i>y</i> be the number of 6 digit numbers, the sum of whose digits is odd, then</p>
<p>(A) <i>x</i> + <i>y</i> = 9 × 10<sup>5</sup></p>
<p>(B) <i>x</i> < <i>y</i></p>
<p>(C) <i>x</i> = <i>y</i></p>
<p>(D) <i>y</i> = 450000</p>
Step-by-Step Solution
Key Concept: The sum of digits is even or odd with equal probability across all 6-digit numbers due to symmetry.
<p><strong>Analysis:</strong> Total 6-digit numbers = 9 × 10<sup>5</sup>. These are divided into two groups: those with even digit sum and those with odd digit sum. By symmetry (toggling the last digit changes parity of sum), exactly half have even sum and half have odd sum.</p><p>∴ <i>x</i> = <i>y</i> = 4.5 × 10<sup>5</sup></p>
Correct Answer: C