Q36 / 100 · BITSAT 2023 (Memory Based Paper 2) · Mathematics

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
  • Correct answer
    C

    60
  • D

    180

Solution

$\mathrm{X}-\mathrm{X}-\mathrm{X}-\mathrm{X}-\mathrm{X}$. 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 no. of arrangements $=6 \times 10=60$ ways

Hence, the correct answer is option: 'C'.

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