Permutations & Combinations
Grade None
Question:
<p>How many numbers greater than hundred and divisible by 5 can be made from the digits 3, 4, 5, 6, if no digit is repeated?</p>
<p style="display:inline">12</p>
<p style="display:inline">6</p>
<p style="display:inline">30</p>
<p style="display:inline">24</p>
Step-by-Step Solution
Key Concept: To solve counting problems with divisibility and magnitude constraints, fix the required units digit first and then sum the permutations for all valid digit-length cases.
<p>Numbers formed from 3, 4, 5, 6 and which are divisible by 5 have ‘5’ fixed in unit’s place<br />
3 Digit Numbers<br />
H T U<br />
x x 5<br />
<sup>3</sup>P<sub>2</sub> ways<br />
= <span class="math-tex">\(\frac{3 !}{1 !}\)</span> = 3 <span class="math-tex">\(\times\)</span> 2<br />
4 Digit Numbers<br />
Th H T U<br />
x x x 5<br />
<sup>3</sup>P<sub>3</sub> ways<br />
= <span class="math-tex">\(\frac{3 !}{0 !}\)</span> = 3 <span class="math-tex">\(\times\)</span> 2<br />
<span class="math-tex">\(\Rightarrow\)</span> Total number of numbers = 6 + 6 = 12</p>
Correct Answer: A