Permutations & Combinations
Grade None
Question:
<p>How many numbers can be formed from the digits 1, 2, 2, 5, 6 which are greater than 50000?</p>
<p style="display:inline">24</p>
<p style="display:inline">30</p>
<p style="display:inline">120</p>
<p style="display:inline">12</p>
Step-by-Step Solution
Key Concept: Identify the valid leading digits that satisfy the numerical constraint and multiply by the permutations of the remaining digits, accounting for identical items using the multiset permutation formula.
<p>The numbers which are greater than 50000 have either 5 or 6 at the ten thousand’s place<br />
<span class="math-tex">\(\Rightarrow\)</span> The ten thousand’s place, can be filled in 2 ways<br />
The remaining 4 digits can be arranged in <span class="math-tex">\(\frac{4 !}{2 !}\)</span> ways (as 2 is repeated)<br />
<span class="math-tex">\(\Rightarrow\)</span> The required number of numbers<br />
= 2 <span class="math-tex">\(\times \frac{4 !}{2 !}\)</span> = 24</p>
Correct Answer: A