Nat 5 questions
- Thread starter mkxc3
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amathproblemthatneedsolve
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- May 12, 2019
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I'd suggest to firstly, divide the lockers into two groups, one for the odd-numbered lockers and the other group for even-numbered lockers. We know that If a color is used in one group, then it can't be used in the other group. Then we can divide into 2 cases.
First case: only two colors are used.
First, we choose two colors, then one of them will be used for the first group and the other will be used for the second group. So the number of ways for this case is P32=6.
Second case: all three colors are used.
We can have two colors in a group and one color for the other group. WLOG, we can assume that red and blue is used for the first group and green for the second group, then multiply the result by C3/2×2=6, then we will have the answer for this case.
We have 2^5−2=30 choices for the first group because there are 2 choices for each of the 5 lockers in the first group, but subtracting 2 for the cases that all of the 5 lockers in the first group are red or blue. Then, there is only 1 case for the second group. That means there are 30×6=180 if you solve this it's the answer to your question
So work out what 180+6=?
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