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

n1n2n3 can be arrranged in 3! ways ryt ??

n1n2n3 can be arrranged in 3! ways ryt ??

2 votes

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 ??

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 ??

1 vote

Best answer

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

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