#combinatorics

Question

In how many different ways can a set A of 3n elements be partitioned into 3 subsets of equal number of elements?

My approach :

The number of permutations of n  objects with n1 identical objects of type 1,  n2 identical objects of type 2, , and n3 identical objects of type  is  n!/n1!n2!n3!

Here ans could be (3n)!/(n!)^3

but given is  3n)!/6* (n!)^3

How did again 6 come in denominator , why is he arranging again in 6 i..e., 3! ways ??

Answer 1

total elements in group  = 3n  , it has to be partitioned / divided  into 3 equal subsets /group
let eeach of three group contains n elements  all of equal sizes
so according to division formula  it will be   $\frac{(3n!)}{(n!)^3 (3!)}$

6 = 3!  came due to power of   n! which is 3

Comments

n!/n1!n2!n3! why no 3! ??

n1n2n3 can be arrranged in 3! ways ryt ??

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If question would   have been :  how many different ways can a set A of 3n elements be partitioned into 3 subsets of unequal number of elements  

then it will be $\frac{(3n!)}{(n!)^{3}}$

since groups  are not of same size 

your doubt :  n!/n1!n2!n3! why no 3! ??

n1n2n3 can be arrranged in 3! ways ryt ??

no need of doing it in question he is not  asking for permutations , only this is we have  3 equal groups  and we have to divide it  

Thank you !!