The total number of six digit numbers, formed using the digits 4, 5, 9 only and divisible by 6, is ____________.
Answer (integer)
81
Solution
A number will be divisible by 6 iff the digit at the unit place of the number is divisible by 2 and sum of all digits of the number is divisible by 3 .
<br/><br/>Units, place must be occupied by 4 and hence, at
least one 4 must be there.
<br/><br/>Possible combination of 4, 5, 9 are as follows :
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.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
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.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
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<table class="tg">
<thead>
<tr>
<th class="tg-0pky">4</th>
<th class="tg-0pky">5</th>
<th class="tg-0pky">9</th>
<th class="tg-0pky">Total number<br> of Number</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-0pky">1</td>
<td class="tg-0pky">1</td>
<td class="tg-0pky">4</td>
<td class="tg-0pky">${{5!} \over {4!}} = 5$</td>
</tr>
<tr>
<td class="tg-0pky">1</td>
<td class="tg-0pky">4</td>
<td class="tg-0pky">1</td>
<td class="tg-0pky">${{5!} \over {4!}} = 5$</td>
</tr>
<tr>
<td class="tg-0pky">2</td>
<td class="tg-0pky">2</td>
<td class="tg-0pky">2</td>
<td class="tg-0pky">${{5!} \over {2!2!}} = 30$</td>
</tr>
<tr>
<td class="tg-0pky">3</td>
<td class="tg-0pky">0</td>
<td class="tg-0pky">3</td>
<td class="tg-0pky">${{5!} \over {2!3!}} = 10$</td>
</tr>
<tr>
<td class="tg-0pky">3</td>
<td class="tg-0pky">3</td>
<td class="tg-0pky">0</td>
<td class="tg-0pky">${{5!} \over {2!3!}} = 10$</td>
</tr>
<tr>
<td class="tg-0pky">4</td>
<td class="tg-0pky">1</td>
<td class="tg-0pky">1</td>
<td class="tg-0pky">${{5!} \over {3!}} = 20$</td>
</tr>
<tr>
<td class="tg-0pky">6</td>
<td class="tg-0pky">0</td>
<td class="tg-0pky">0</td>
<td class="tg-0pky">${{5!} \over {5!}} = 1$</td>
</tr>
</tbody>
</table>
<br/><br/>Total = 5 + 5 + 30 + 10 + 10 + 20 + 1 = 81.
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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