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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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 $\frac{4!}{2! 2!} = 6$ ways. The five digits 2, 2, 8, 8, 8 can be arranged at (X) place in $\frac{5!}{2! 3!} = 10$ ways.
Total numbers of arrangement 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 ?

The correct option is C 60X - X - X - X - X The four digits 3, 3, 5, 5 can be arranged at (-) places in 4!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 numbers of arrangement is 6×10=60.

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