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

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

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 == Solution ==
-{{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 ==

2008 UNCO Math Contest II (Problems • Answer Key • Resources)
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

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

Revision as of 01:04, 13 January 2019

Problem

Solution

See Also