Question

How many different nine-digit numbers can be formed from the number $22 \: 33 \: 55 \: 888 $ by rearranging its digits so that the odd digits occupy even positions?

• A. 16
• B. 36
• C. 60
• D. 180

Answer 1

Answer: (C)

$X - X - X - X - X.$ The four digits $3, 3, 5, 5$ can be arranged at (- ) places in $\frac{4!}{2!2!}=6$ways
The five digits $2, 2, 8, 8, 8 $ can be arranged at $(X)$ places in $\frac{5!}{2! 3!}ways=10 \, ways$
Total number of arrangements $= 6 \times 10 = 60$
[since, events $A$ and $B$ are independent, therefore
$ A\cap B=A\times B]$