Permutations & Combinations
Grade None

Question:

<p>Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is:</p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{2}\)</span>(6!)</p>
<p style="display:inline"><span class="math-tex">\(\frac{5}{2}\)</span>(6!)</p>
<p style="display:inline">5<sup>6</sup></p>
<p style="display:inline">6!</p>

Step-by-Step Solution

Key Concept: To form a 6-digit number using all five specified digits, exactly one digit must be repeated, requiring a combination of digit selection and multiset permutations.
<p>Five digits numbers be 1, 3, 5, 7, 9<br /> For the selection of one digit, we have <sup>5</sup>C<sub>1</sub>&nbsp;choice.<br /> And six digits can be arranged in&nbsp;<span class="math-tex">$\frac{6 !}{2 !}$</span>&nbsp;ways.<br /> Hence, total such numbers =&nbsp;<span class="math-tex">$\frac{5.6 !}{2 !}=\frac{5}{2}$</span>6!</p>
Correct Answer: B

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