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Permutation and Combination Problem

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Permutation and Combination Problem

by sukhman » Mon Sep 09, 2013 10:36 pm
A nickel, a dime, and two identical quarters are arranged along a side of a table. If the quarters and the dime have to face heads up, while the nickel can face either heads up or tails up, how many different arrangements of coins are possible? (4! / 2!) × 2!
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by ganeshrkamath » Tue Sep 10, 2013 12:03 am
sukhman wrote:A nickel, a dime, and two identical quarters are arranged along a side of a table. If the quarters and the dime have to face heads up, while the nickel can face either heads up or tails up, how many different arrangements of coins are possible? (4! / 2!) × 2!
Number of ways of arranging a nickel, a dime, and two indentical quarters
= 4!/2! (divided by 2! to account for the identical quarters)

Now, since the nickel can either be facing heads up or tails up, the total number of ways = (4!/2!) * 2

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