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Let {3n!}{k(n!)^3} be the number of different ways a set A of

Question

Let $\frac{3 n !}{k ( n ! ) ^{3}}$ be the number of different ways a set $A$ of $3 n$ elements be partitioned into $3$ subsets of equal number of elements? (the subsets $P, Q, R$ form a partition if $P \cup Q \cup R = A, P \cap R = ϕ, Q \cap R = ϕ, R \cap P = ϕ) .$ Find $k$ ?

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Let

\frac{3 n !}{k ( n ! ) ^{3}}

be the number of different ways a set

A

3 n

elements be partitioned into

3

subsets of equal number of elements? (the subsets

P, Q, R

form a partition if

P \cup Q \cup R = A, P \cap R = ϕ, Q \cap R = ϕ, R \cap P = ϕ) .

Find

k

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

The required number of ways

=

the number of ways in which

3 n

different things can be divided in

3

equal groups

=

The number of ways to distribute

3 n

different things equally among three persons

= \frac{3 n !}{3 ! ( n ! ) ^{3}}

= \frac{3 n !}{6 ( n ! ) ^{3}}

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Question Text	Let $\frac{3 n !}{k ( n ! ) ^{3}}$ be the number of different ways a set $A$ of $3 n$ elements be partitioned into $3$ subsets of equal number of elements? (the subsets $P, Q, R$ form a partition if $P \cup Q \cup R = A, P \cap R = ϕ, Q \cap R = ϕ, R \cap P = ϕ) .$ Find $k$ ?
Answer Type	Text solution:1
Upvotes	150