Difference between revisions of "2008 UNCO Math Contest II Problems/Problem 10"

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-== Problem ==
-Let <math>f(n,2)</math> be the number of ways of splitting <math>2n</math> people into <math>n</math> groups, each of size <math>2</math>. As an example,
-the <math>4</math> people <math>A, B, C, D</math> can be split into <math>3</math> groups: <math>\fbox{AB} \ \fbox{CD} ; \fbox{AC} \ \fbox{BD} ;</math>
-and <math>\fbox{AD} \ \fbox{BC}.</math>
-Hence <math>f(2,2)= 3.</math>
-(a) Compute <math>f(3,2)</math> and <math>f(4,2).</math>
-(b) Conjecture a formula for <math>f(n,2).</math>
-(c) Let <math>f(n,3)</math> be the number of ways of splitting <math>\left \{1, 2, 3,\ldots ,3n \right \}</math> into <math>n</math> subsets of size <math>3</math>.
-Compute <math>f(2,3),f(3,3)</math> and conjecture a formula for <math>f(n,3).</math>
-== Solution ==
-(a) <math>f(3,2)=15</math>
-(b) <math>f(n,2)=(2n-1)!!</math>
-(c) <math>f(2,3)=\binom{5}{2} ; f(3,3)=\binom{8}{2} \binom{5}{2} ;f(n,3)=\frac{(3n)!}{6^n \cdot n!} </math>
-== See Also ==
-{{UNCO Math Contest box|n=II|year=2008|num-b=9|after=Last Question}}
-[[Category:Intermediate Combinatorics Problems]]

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