Bifalgorithm post · Brisk · Chapter 1

What are the three basic situations behind most counting problems?

What are the three basic situations behind most counting problems? (Brisk · Chapter 1) — "Most of the time, counting problems all boil down to a variation of o…

Ideas in this flowchart

  • Most counting problems reduce to one of three basic situations.
  • The three basic counting situations
  • n^k: k different things each have n completely independent choices.
  • n!/(n-k)!: choose an ordered sequence of k different things from n without repeats.
  • n!: the special case of n!/(n-k)! when k = n.
  • (n choose k): choose k different things from n when order does not matter.
  • To identify the applicable situation, analyze the counting problem carefully.
  • Ask whether repeated values are allowed.
  • Ask whether the order of the values matters.
  • Determining which situation applies is often subtle.
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What are the three basic situations behind most counting problems?

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Brisk

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"Most of the time, counting problems all boil down to a variation of one of the following three basic situations: • n^k — this is when we have k different things, each of which is free to take on one of n completely independent choices. • n!/(n-k)! — this is when we’re taking a sequence of k different things from a set of n, but no repeats are allowed. (A special case of this is n!, when k = n.) • (n choose k) — this is when we’re taking k different things from a set of n, but the order doesn’t matter. Sometimes it’s tricky to deduce exactly which of these three situations apply. You have to think carefully about the problem, and ask yourself whether repeated values would be allowed, and whether it matters what order the values appear in. This is often subtle."

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