Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is :
Solution
Here none number repeats more than once.
<br><br>We can choose the number which repeats more than once among 1, 3, 5, 7, 9 in <sup>5</sup>C<sub>1</sub> ways.
<br><br>Let number 3 repeats more than once. So six digits are 1, 3, 3, 5, 7, 9.
<br><br>We can arrange those six digits in ${{6!} \over {2!}}$ ways.
<br><br>$\therefore$ Total six digit numbers = <sup>5</sup>C<sub>1</sub> $\times$ ${{6!} \over {2!}}$ = ${5 \over 2}\left( {6!} \right)$
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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