Exercise · ZW-STIRLING-NUMBER-OF-THE-FIRST-KIND-010 · ★★★ Advanced

A claim says \(c(n,k)=c(n-1,k-1)+c(n-1,k)\) always holds. For which pair \((n,k)\) does the equality actually hold?

Choose one option.

Show answer
  • B. \((2,1)\)

Explanation

For \((2,1)\) both sides equal 1. For \((4,1)\), \((3,1)\) and \((3,2)\) the right side gives 2, 1 and 2, while \(c(n,k)\) equals 6, 2 and 3.

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Review the concept: Stirling number of the first kind

The Stirling number of the first kind \(c(n,k)\) counts the permutations of \(n\) elements that have exactly \(k\) cycles.

\[x(x+1)\cdots(x+n-1)=\sum_{k=0}^{n} c(n,k)\,x^{k}\]

More practice problems: stirling number of the first kind

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