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Q. 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

Permutations and Combinations

Solution:

Total number of ways $=6 \times 6 \times \ldots$ to $n$ times $=6^{n}$
Total number of ways to show only even numbers $=3 \times 3 \times\ldots n$ times $-3^{n}$
Therefore, the required number of ways $=6^{n}-3^{n}$