I have three pairs of shoes. I plan to wear them over the next ten days as follows: I dont want to wear the same pair on two consecutive days, and I dont want to wear the same pair on the first day and the tenth. how many combinations are there?

Plase note: I do not have to wear all the pairs. Also, the formula should hold true regardless of the amount of days OR the amount of shoes (the first day cannot be the same as the LAST, not necessarily the tenth.)

Where did you get this formula from?
ago
reopened ago
I'm curious to know:

1) the logic behind this formula

2) who was the first person to figure this out

Dear GeniusBY

Just to let you know I saw your comment and private message. I’m pleased you want to see how the formula was derived. I didn’t look it up but derived it from logic and observation by working out the possible combinations up to 5 days. I built a table and spotted a pattern which gave me the formula. It will take a day or two to write it all in a comment, and if you live in the USA, your time will be 5 or 6 hours behind U.K. time. Watch this space!

So you have a table already. I inserted 2 columns. Of the combinations, I noted how many ended with the same pair of shoes as Day 1 after n days and how many ended with a different pair. The sum is 3(2ⁿ⁻¹), the total number of combinations. The number in the “same shoes” column for Day n becomes the number in the “different shoes” column for the next day, Day n+1. So for Day n+1 it’s easy to calculate the “same shoes” figure for Day n+1. A quick inspection shows that the figures are a power of 2 plus or minus 2. It’s minus when n is odd and plus when n is even. That brings in the term 2(-1)ⁿ. The reason I think this works is that when we end up with a different pair of shoes, the next day will always include the same pair of shoes because we can’t have the same shoes on consecutive days. For example, if Day 1 was shoes A, then Day n would be be A for “same shoes” and B or C for “different shoes”. B can only be followed the next day by A or C, and C can only be followed by B or A, so there will always be a “same shoes” count for the next day. This fact enables the table to be completed.

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Best answer

Finding a formula is not the way to look at the problem. That’s the lazy approach! If we had to have a formula for every problem we would have so many formulae we would need a formula to find the one we’re looking for! It’s all about logical thinking. Sometimes a formula will arise when we think through a problem logically. But the big clue is to look for a pattern. Formulae often arise when we spot a pattern.

Let’s see if a pattern comes out of this problem.

Let’s go through each day. On Day 1, we have a choice of 3 pairs of shoes. Label the pairs A, B, C.

On Day 2 we can have AB, AC, BA, BC, CA, CB. That’s 6 combinations, twice as many as Day 1.

On Day 3 we can have ABA, ABC, ACA, ACB, BAB, BAC, BCA, BCB, CAB, CAC, CBA, CBC—12 combinations.

On Day 4 we have twice as many again—24.

Now we can see at least one simple pattern in the number of combinations for the number of days: 3, 6, 12, 24, 48, ..., because each combination “spawns” 2 for the next day. We can write this as a formula: 2ⁿ⁻¹×3 where n is the number of days. So when n=10 there are 2⁹×3=512×3=1536 combinations—but not all will be valid.

I note that you edited the question. I found a formula based on observation which seems to fit the facts and enables us to predict the result for any number of days: 2ⁿ+2(-1)ⁿ. So when n=10, this comes to 1026. I think I can show the logic leading to the formula, if you need it.

ago by Top Rated User (597k points)

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