search
Log In
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
2 votes
121 views
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 ??
in Mathematical Logic 121 views

1 Answer

1 vote
 
Best answer

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


selected by
0
n!/n1!n2!n3! why no 3! ??

n1n2n3 can be arrranged in 3! ways ryt ??
1

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 !!

Related questions

0 votes
2 answers
1
140 views
#COMB There are $4$ boys and $6$ prizes are to be distributed among them such that each has at least $1$ prize. How many ways that can be done? My solution: $\text{Case 1 : 3 1 1 1}$ $\text{Case 2 : 2 2 1 11}$ ... My doubt is in the second case, am I not considering the prizes to be indistinguishable? I am confused in this regard. Please help me clear this doubt.
asked May 25, 2018 in Combinatory Abhisek Das 140 views
1 vote
1 answer
2
526 views
The number of ways in which n distinct objects can be put into two identical boxes so that no box remains empty, is a) 2^n - 1 b) 2^n - 2 c) 2^(n-1) - 1 d) None of these Please explain your answer.
asked Jun 11, 2017 in Combinatory Jatin18 526 views
0 votes
1 answer
3
1 vote
1 answer
4
364 views
At any time, the total number of persons on earth who have shaken hands an odd number of times has to be The answer provided is even number but cannot understand how
asked Sep 23, 2017 in Mathematical Logic VIKAS PAREEK 1 364 views
...