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?

Question

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? (A) 16 (B) 36 (C) 60 (D) 180

Tardigrade Admin · Accepted Answer

X - 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 number of arrangements is 6 × 10=60.

Q. 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

36

60

180

$X - X - X - X - X$
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 number of arrangements is $6 \times 10 = 60$ .

Q. 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

36

60

180

X−X−X−X−X The four digits 3,3,5,5 can be arranged at (−) places in 2!2!4!​=6 ways. The five digits 2,2,8,8,8 can be arranged at (X) place in 2!3!5!​=10 ways. Total number of arrangements is 6×10=60.

Q. 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?

$X - X - X - X - X$
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 number of arrangements is $6 \times 10 = 60$ .