# combinatorial problem

• Apr 18th 2010, 10:47 AM
baz
combinatorial problem
four persons,A,B,C,D form a committee.One is to be president,one is to secretary, one is to be finance secretary and one should be registrar. In how many ways can these posts be asigned?

But how can this bepossible as as there will be many ways in which each person will have more than one post,and in some ways all the posts will go to one person.
Then how to solve this problem.
• Apr 18th 2010, 11:36 AM
Quote:

Originally Posted by baz
four persons, A,B,C,D form a committee. One is to be president, one is to be secretary, one is to be finance secretary and one should be registrar. In how many ways can these posts be assigned?

But how can this bepossible as as there will be many ways in which each person will have more than one post,and in some ways all the posts will go to one person.
Then how to solve this problem.

The four people form a committee.
Hence each one will occupy one post exclusively.

If one person could perform up to all roles, the answer would be $\displaystyle 4^4$

as there would always be 4 ways to fill each position.

The factorial version bases it's result on the fact that each person assumes only one role.
• Apr 18th 2010, 12:04 PM
baz
4! is number of ways of arranging 4 positions, but is there any way of knowning what exactly are these permutations?
In my view, the basis of the answer(4!) is on castersian product of n sets and this does not give actual permutations.
• Apr 18th 2010, 12:04 PM
CausingTraumaZ
You should use variations here because here it's important the position: It's not the same to be secretary and registrar.
The formula for variations is

$\displaystyle V=\frac{n!}{(n-k)!}\$

where n is the number of elements and k is the number of classes.
• Apr 18th 2010, 12:07 PM
baz
Quote:

Originally Posted by CausingTraumaZ
You should use variations here because here it's important the position: It's not the same to secretary and registrar.
The formula for variations is

$\displaystyle V=\frac{n!}{(n-k)!}\$

where n is the number of elements and k is the number of classes.

I think we cannot use this aproach as this will goes us set of ordered triplets,so correct answer is 4!.
• Apr 18th 2010, 12:22 PM
Quote:

Originally Posted by baz
4! is number of ways of arranging 4 positions, but is there any way of knowning what exactly are these permutations?
In my view, the basis of the answer(4!) is on castersian product of n sets and this does not give actual permutations.

As we arrange the 4 people among the 4 positions,
or arrange the 4 positions among the 4 people,
we can class it as a permutation

$\displaystyle 4p_4$ if desired, where we arrange all 4 elements of the set

in all possible arrangements, which is $\displaystyle \frac{4!}{(4-4)!}$

As this amounts to 4 choices for position one,
for each of these choices there are 3 remaining choices for position two,
for each of those 3 there are 2 choices for position three,

which is $\displaystyle 4!$

$\displaystyle 0!$ has to be equated to 1.

By listing the possibilities instead, the pattern is apparent.