Exercise · ZW-STIRLING-NUMBER-OF-THE-SECOND-KIND-006 · ★★☆ Intermediate
Use the recurrence to compute S(4,2).
Choose one option.
Show answer
- B. 7
Explanation
Applying the recurrence, S(4,2)=2S(3,2)+S(3,1)=2·3+1=7. The value 6 comes from dropping the S(3,1) term in the sum.
Review the concept: Stirling number of the second kind
The Stirling number of the second kind S(n,k) counts the partitions of a set of n elements into k nonempty, indistinguishable blocks.
\[S(n,k)=kS(n-1,k)+S(n-1,k-1)\]
More practice problems: stirling number of the second kind
- ZW-STIRLING-NUMBER-OF-THE-SECOND-KIND-001 How many partitions of a 3-element set into 2 nonempty blocks exist?
- ZW-STIRLING-NUMBER-OF-THE-SECOND-KIND-002 State the value of S(n,n) for any n≥1.
- ZW-STIRLING-NUMBER-OF-THE-SECOND-KIND-003 State what the Stirling number S(n,k) counts.
- ZW-STIRLING-NUMBER-OF-THE-SECOND-KIND-004 What is the value of S(n,1) for n≥ 1?
- ZW-STIRLING-NUMBER-OF-THE-SECOND-KIND-005 What is S(n,k) when k>n?
- ZW-STIRLING-NUMBER-OF-THE-SECOND-KIND-007 Compute S(4,3) using S(n,k)=kS(n-1,k)+S(n-1,k-1) and S(3,3)=1.
- ZW-STIRLING-NUMBER-OF-THE-SECOND-KIND-008 Add the values S(3,k) for k=1,2,3 and state which Bell number this is.
- ZW-STIRLING-NUMBER-OF-THE-SECOND-KIND-009 How many ordered partitions of 3 elements into 2 nonempty blocks exist?
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