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In how many different ways can a nickel, a dime and a quarter be given to four children if one coin is given to each child

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In this case, we have to use permutations to solve this problem, we have 3 options for 3 positions, use permutation this way:


\begin{gathered} 3P3=(3!)/((3-3)!) \\ 3P3=(3!)/(0!)=6 \end{gathered}

The coins can be given in 6 different ways.

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