getso wrote:Hi,
Can anyone explain how did we arrive at 3 !. I understood the first part.
Regards,
Shobha
stop@800 wrote:First person can be selected in 8 ways [can be any one]
second person can be selected in 6 ways [keep aside the spouse of person just selected]
thrd person can be selected in 4 ways [keep aside the spouse of two persons selected]
total 8 * 6 * 4
now our selection is order dependent
so we need to divide it by 3! to make it indep of order
Remember ABC and BAC are same committee
has this been some prize and we had first second and third prize
than order is imp
so we do not need to divide by 3!
8 * 6 * 4 / 3! = 32
There is another method also using combinations
but I refrained from that as it may confuse you
Let's say you had a pair of objects: A and B. If order doesn't matter, then how many pairs do you have? Clearly AB is
one pair. And if order doesn't matter, then it doesn't matter whether you call that pair AB or BA--it's still one pair--or one "unordered" group of two objects. But if order matters, then AB is one "ordered" pair and BA is another ordered pair. So, notice that when order matters we have 2 pairs, and when order doesn't matter it's just 1 pair. To go from the number of ordered groups to the number of unordered groups, you have to divide by the factorial of the number of objects. A and B are two objects. There are two ordered pairs. To remove order: 2/2! = 1 unordered pair.
Let's say you have three objects: A, B and C. That's
one unorderered group of three. But there are 3! or
six ordered groups (number of ways of arranging n objects is n!). To go from from the number of ordered groups to the number of unordered groups, we would just divide the number of ordered groups--6--by the factorial of the number of objects--3!, giving us 1.
Now, in the above two paragraphs I was talking about groups. But this problem and GMAT combinatorics frequently involves selecting subgroups from a larger group. The selection of unordered subgroups are combinations; the selection of ordered subgroups are permutations. Let's consider the formula for selecting combinations and permutations from bigger sets. Let's first look at the permutations formula:
nPk = n!/(n -k)!
in which n is the number of objects in the big group and k is the number of objects in the subgroup. This formula yields the total number of all the
ordered subgroups of size k selected from a bigger group of size n
Now, let's look at the formula for combinations:
nCk = n!/[k!*(n-k)!]
This formula yields the total number of all the
unordered subgroups of size k selected from a bigger group of size n. So, notice that the only difference in the two formulae is that in the combinations formula we are, again, dividing by the factorial of the number of objects in the subgroup. So, the only difference between the two formulae acknowledges that when we are unconcerend about order, we must divide by the factorial of the number of objects in the group or subgroup.
In this problem, when we place any of the eight people in the first slot, leaving six for the second, and four for the third, ABC will end up being counted as a different committee from BCA. But because they are comprised of the same people, ABC is the same committee as BAC, no matter whether you call them "Andy, Bob, and Charlie" or "Bob, Andy, and Charlie". In other words, in this problem, order does not matter. This is why we have to divide 8*6*4 by the factorial of the number of objects in the subgroup (committee)--3!.
In combinatorics problems, you should always determine very early on whether order matters or not. In order for order to matter, the problem has to communicate a situation where order would matter: for example, ranking or a code, or using the word "order" or "arrange", etc.
