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1) Six children, Arya, Betsy, Chen, Daniel, Emily and Franco are to be seated in a single row of six chairs. If Betsy cannot sit next to Emily, how many different arrangements of the six children are possible?

So the total number of arrangements is 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720

There are 5 ways that Betsy is to the left of Emily and there are 5 ways that Emily is to the left of Betsy. So there are 10 ways total that Emily and Betsy violate the question's parameters. So, for any of those 10 ways, the 4 remaining children can be seated in 4! ways: 4! = 4 x 3 x 2 x 1 = 24. So you need to subtract 24 x 10 = 240 ways from the original total of 720. So the answer is 480.

Generally speaking, the method in this question is total combos - combinations that violate.

2) If a coin is flipped 5 times, what is the probability of exactly 3 heads?

So HHHTT is 1/32 but that's only 1 particular order. We also need to count HHTTH. We need to multiply 1/32 by the number of ways that we can get this outcome.

The formula for this problem is going to be 5! / (3!2!) = 10. We calculate the combinations this way because we need to divide out the indistinguishable outcomes from each other. For example H1H2H3T1T2 = H3H2H1T1T2 = H2H3H1T1T2 etc. Those 3 heads in the front can be arranged in 3! ways and so we need to divide out those from the total.

Anyway, the answer to this problem is 1/32 x 10 = 5/16.

So why the difference in approach? Why are we dividing out the combinations that are duplicates in question 2 but we're subtracting out the duplicates in question 1? I can't wrap my head around it! What's going on?

Jwan622
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3 Answers3

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The method in (1) is rather strange. We sit Betsy, and Emily to her right. There are five places the pair of them can sit. Then the four remaining seats need to be filled, in $4!$ ways.

What would be more natural in my mind is to consider $[BE]$ to be a single unit, and we permute the five letters $A, [BE], C, D, F$, which can be done in $5!=120$ ways.

Problem 2 is different because one H is just like any other, but the five people are distinguishable. If among the people we had two identical twins that we couldn't tell apart, we would be dividing the answer by 2 in the end.

vadim123
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  • I think we also need to account for the fact that Betsy can be on the right too and Emily on the left. So, there are 10 places the pair can be next to each other. – Jwan622 Nov 21 '13 at 18:09
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My initial thoughts are that the answer lies in the fact that in question 1) the people are all different whereas in question 2) all the heads are identical and all the tails are identical... but i'm not sure if this is what's causing the difference.

Like, in problem 1, we still have to calculate the combinations of the other 4 after determining that there are 10 combinations of Emily and Betsy that violate. This is because in question 1, we don't have a duplication problem like in question 2 where we take all the duplicate combinations and reduce them to 1. In question 1, we actually have to remove the unwanted combinations entirely, not just singly count them. This could be the source of the different methods right?

Is there a way to solve problem 1 with division and not subtraction? How would it have to be worded?

Jwan622
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The difference is that in the first question all arrangements are unique, and we are subtracting particular arrangements that are not allowed. There are no duplicates to deal with.

In the second question we are instead counting every individual arrangement $3!2!$ times, and the division corrects this.

Jonas Granholm
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