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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Permutation and Combination

BITSAT 2020 · Mathematics · Permutation and Combination

The number of possible outcomes in a throw of n ordinary dice in...

The number of possible outcomes in a throw of n ordinary dice in which at least one of the dice shows an odd number is

Options

  1. A6n−16^{n}-1
  2. B3n−13^{n}-1
  3. C6n−3n6^{n}-3^{n}
  4. DNone of these

Correct answer

C. 6n−3n6^{n}-3^{n}

Step-by-step solution

Total number of ways =6×6×……=6 \times 6 \times \ldots \ldots to nn times =6n=6^{n} Total number of ways to show only even number =3×3×……=3 \times 3 \times \ldots \ldots to nn times =3n∴=3^{n}\therefore required number of ways =6n−3n=6^{\mathrm{n}}-3^{\mathrm{n}}
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