Question

Total number of six-digit numbers in which only and all the five digits $1,3,5,7$ and $9$ appear, is

• A. $5^6$
• B. $\frac{1}{2}(6!)$
• C. $6!$
• D. $\frac{5}{2}(6!)$

Answer 1

The correct option is D $\frac{5}{2}(6!)$

Selecting all $5$ digits $={}^5{C}_5=1$ way
Now, we need to select one more digit from $1,3,5,7,9$ to make it a $6$-digit number.
This can be done in ${}^5{C}_1=5$ ways
Total number of permutations $=\frac{6!}{2!}$
Therefore, total numbers $={}^5{C}_5\times {}^5{C}_1\times \frac{6!}{2!}=\frac{5}{2}(6!)$