How many different nine digit numbers can be formed from the number $223355888$ by rearranging its digits so that the odd digits occupy even positions?

16

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36

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60

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180

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Solution

The correct option is C $60$
$X - X - X - X - X$
The four digits $3, 3, 5, 5$ can be arranged at $(-)$ places in
$\frac{4!}{2! 2!} = 6 w a y s$ . The five digits $2, 2, 8, 8, 8$ can be arranged at $(X)$ place in $\frac{5!}{2! 3!} = 10 w a y s$ .
Total number of arrangements is $6 \times 10 = 60$ .

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How many different nine- digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions?

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How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions?

The correct option is C 60X−X−X−X−XThe four digits 3,3,5,5 can be arranged at (−) places in4!2!2!=6 ways. The five digits 2,2,8,8,8 can be arranged at (X) place in 5!2!3!=10 ways.Total number of arrangements is 6×10=60.

How many different nine digit numbers can be formed from the number $223355888$ by rearranging its digits so that the odd digits occupy even positions?

The correct option is C $60$
$X - X - X - X - X$
The four digits $3, 3, 5, 5$ can be arranged at $(-)$ places in
$\frac{4!}{2! 2!} = 6 w a y s$ . The five digits $2, 2, 8, 8, 8$ can be arranged at $(X)$ place in $\frac{5!}{2! 3!} = 10 w a y s$ .
Total number of arrangements is $6 \times 10 = 60$ .