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

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== Solution ==
 
== Solution ==
{{solution}}
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(a) <math>f(3,2)=15</math> 
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(b) <math>f(n,2)=(2n-1)!!</math>
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(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 ==
 
== See Also ==

Revision as of 01:04, 13 January 2019

Problem

Let $f(n,2)$ be the number of ways of splitting $2n$ people into $n$ groups, each of size $2$. As an example,

the $4$ people $A, B, C, D$ can be split into $3$ groups: $\fbox{AB} \ \fbox{CD} ; \fbox{AC} \ \fbox{BD} ;$ and $\fbox{AD} \ \fbox{BC}.$

Hence $f(2,2)= 3.$

(a) Compute $f(3,2)$ and $f(4,2).$

(b) Conjecture a formula for $f(n,2).$

(c) Let $f(n,3)$ be the number of ways of splitting $\left \{1, 2, 3,\ldots ,3n \right \}$ into $n$ subsets of size $3$. Compute $f(2,3),f(3,3)$ and conjecture a formula for $f(n,3).$


Solution

(a) $f(3,2)=15$

(b) $f(n,2)=(2n-1)!!$

(c) $f(2,3)=\binom{5}{2} ; f(3,3)=\binom{8}{2} \binom{5}{2} ;f(n,3)=\frac{(3n)!}{6^n \cdot n!}$

See Also

2008 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Last Question
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions