Problem 4 · AMC 8 Stretch
Stretch
Logic & Word Problems
Counting & Probability
account-for-all-possibilitiesorganizing-datareduce-and-expandchoosing-vs-arranging
A captured hero must design a new flag with three vertical stripes, each stripe a different color. He can only use colors that already appear on these five real three-stripe flags, which together use \(6\) different colors: Belgium (black, yellow, red); France (blue, white, red); Ireland (green, white, yellow); Italy (green, white, red); Romania (blue, yellow, red). How many different three-stripe flags can be made from these \(6\) colors? How many of those are brand new (not one of the five that already exist)?
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Answer: 120 possible flags in all; 115 of them are new
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Hint 1 of 4
There are really two questions hiding here: WHICH three colors to use, and in WHAT order to put them left to right. Solve them one at a time.
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Hint 2 of 4
First count how many ways to choose \(3\) different colors out of \(6\) when order doesn't matter yet. List them carefully so you don't miss any.
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Hint 3 of 4
There are \(20\) ways to choose the three colors. Now take one set of three colors and count the left-to-right orders: ABC, ACB, BAC, BCA, CAB, CBA. How many is that?
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Approach: Separate choosing the colors from arranging them, then subtract existing flags
- Step 1 — choose the three colors: pick 3 of the 6 with order not mattering yet; there are exactly 20 sets (\(\binom{6}{3}=\dfrac{6\times5\times4}{3\times2\times1}=20\)).
- Step 2 — order each set: three chosen colors A, B, C arrange left-to-right as ABC, ACB, BAC, BCA, CAB, CBA — exactly 6 (\(3!=6\)). Since stripes are ordered, each order is a different flag.
- Step 3 — total flags: \(20\times6=120\). (Directly: 6 choices for the first stripe, 5 for the second, 4 for the third, \(6\times5\times4=120\).)
- Step 4 — new flags: remove the 5 that already exist: \(120-5=115\) brand-new flags.
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