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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Mathematics
Chapter
Permutation and Combination

BITSAT 2023 · Paper 2 · Mathematics · Permutation and Combination

How many different nine digit numbers can be formed from the number...

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?

Options

  1. A
    16
  2. B
    36
  3. C
    60
  4. D
    180

Correct answer

C.
60

Step-by-step solution

X−X−X−X−X\mathrm{X}-\mathrm{X}-\mathrm{X}-\mathrm{X}-\mathrm{X}. The four digits 3, 3, 5,5 can be arranged at (−)(-) places in 4!2!2!=6\frac{4 !}{2 ! 2 !}=6 ways. The five digits 2,2,8,8,82,2,8,8,8 can be arranged at (X)(X) places in 5!2!3!\frac{5 !}{2 ! 3 !} ways =10=10 ways
Total no. of arrangements =6×10=60=6 \times 10=60 ways
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