Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

there is probably some formula for this - but i never remember formulas and always get confused in choosing which one to use. so i alwaus count, and use simple reasoning. here is how i did it.

a) how many ways there are to divide 4 people into two group of 2?
that's simple counting: (ab,cd) (ac,bd) (ad,bc) - total 3.
b) how many ways there are to divide 6 people into 3 groups of 2?
there are 5 possibilities for member a:
(ab) (ac) (ad) (ae) (af)
for each such possibility we are left with 4 poeple that need to be grouped into 2 groups of 2. but we already did that (!!!!!!) so for each of the 5 options we have 3 possible completions. total of 5*3=15 options.

c) how many ways to divide 8 people into 4 teams of 2?
use the same reasoning: for person a there are 7 possibilities. and for each such possibility we are eft with dividing 6 people to 3 pairs, which we laready did. hence the answer is 7*(5*3) = 105

with this reasoning (if you followed it fully....) you can answer in no time how many ways there are to divide 12 people into 6 pairs, or 20 people into 10 pairs...

there is probably some formula for this - but i never remember formulas and always get confused in choosing which one to use. so i alwaus count, and use simple reasoning. here is how i did it.

a) how many ways there are to divide 4 people into two group of 2? that's simple counting: (ab,cd) (ac,bd) (ad,bc) - total 3. b) how many ways there are to divide 6 people into 3 groups of 2? there are 5 possibilities for member a: (ab) (ac) (ad) (ae) (af) for each such possibility we are left with 4 poeple that need to be grouped into 2 groups of 2. but we already did that (!!!!!!) so for each of the 5 options we have 3 possible completions. total of 5*3=15 options.

c) how many ways to divide 8 people into 4 teams of 2? use the same reasoning: for person a there are 7 possibilities. and for each such possibility we are eft with dividing 6 people to 3 pairs, which we laready did. hence the answer is 7*(5*3) = 105

with this reasoning (if you followed it fully....) you can answer in no time how many ways there are to divide 12 people into 6 pairs, or 20 people into 10 pairs...

Boy! what a clear thinking you have !!! Awesome solution

Hi Hobbit,
I am sorry, I am not able to understand your reasoning behind that.
When you say dividing 4 people into 2 groups of 2 each, I got the following # of ways :

I think we need to go one more step ahead.
Consider the individual members as a, b, c, d, e, f, g, and h

Here, the last group 2 C 2 = 1. Say it is g and h.

Now, try to visualize ...
Havenâ€™t we considered this same group while calculating 8 C 2. Yes.
8 C 2 included this group, 6 C 2 included the same group and 4 C 2 did the same thing...
So g-h group was repeated 4 times.

In the same way, the group under 4 C 2 was repeated 3 times.
In the same way, the group under 6 C 2 was repeated 2 times.
In the same way, the group under 8 C 2 was repeated 1 time.

Actual value is = 8C2x6C2x4C2x2C2 / (4 * 3 * 2 *1) = 2520 / 24 = 105

lets order the8 people and divide into pairs in this way: first and second persons in our ordering are the first pair, third and fourth are another pair and so forth... now if we are going to go over all possible orderings of 8 people we sure going to count all the possible pairings. HOWEVER, there are pairings which are counted more than once. let's se how many times each pairing is counted when going over all possible orderings:

there are total of 8! orderings.

the order in the first pair doesn't matter. so orderings that start with x,y,...
and y,x,... create the same pairing. so we need to divide by 2.
similarly the ordering in the second pair (and the third and the fourth) doesn't matter as well, so we need to divide by 2 for each pair.
in total we have to divide the orderings by 2*2*2*2=16
but that's not the end of it.
the order of the pairs doesn't matter as well. i.e. it doesn't matter if x and y are paired first or second etc...
there are 4!=24 possible ordering of 4 pairs and we need to divide it as well.

so at the end: number of ways = 8!/(2*2*2*2*4!) = 105

this way of counting is also robust, but in my view, is more error prone, as unless you are very experienced and self confident you can always "miscount" - i.e forget to divide something (that is counting the same result more than once), or forget to count a valid solution.

Well done, Hobbit, thatÂ´s exactly how I did it too (no kidding )

You just use recursive reasoning, starting with the smallest number of items, 2 in this case. Then expand progressively to 4, 6, etc, till you see a pattern or finish counting, whichever happens first.

A formula for these problems would be, given n = number of items (people, etc):

(n-1)*(n-3)*(n-5)...*1

In this case: 7*5*3*1 = 105 (B).

Now on for a beer to cool my brain

hobbit wrote:

answer is 105 (B)

there is probably some formula for this - but i never remember formulas and always get confused in choosing which one to use. so i alwaus count, and use simple reasoning. here is how i did it.

a) how many ways there are to divide 4 people into two group of 2? that's simple counting: (ab,cd) (ac,bd) (ad,bc) - total 3. b) how many ways there are to divide 6 people into 3 groups of 2? there are 5 possibilities for member a: (ab) (ac) (ad) (ae) (af) for each such possibility we are left with 4 poeple that need to be grouped into 2 groups of 2. but we already did that (!!!!!!) so for each of the 5 options we have 3 possible completions. total of 5*3=15 options.

c) how many ways to divide 8 people into 4 teams of 2? use the same reasoning: for person a there are 7 possibilities. and for each such possibility we are eft with dividing 6 people to 3 pairs, which we laready did. hence the answer is 7*(5*3) = 105

with this reasoning (if you followed it fully....) you can answer in no time how many ways there are to divide 12 people into 6 pairs, or 20 people into 10 pairs...

Well done, Hobbit, thatÂ´s exactly how I did it too (no kidding )

You just use recursive reasoning, starting with the smallest number of items, 2 in this case. Then expand progressively to 4, 6, etc, till you see a pattern or finish counting, whichever happens first.

A formula for these problems would be, given n = number of items (people, etc):

(n-1)*(n-3)*(n-5)...*1

In this case: 7*5*3*1 = 105 (B).

Now on for a beer to cool my brain

hobbit wrote:

answer is 105 (B)

there is probably some formula for this - but i never remember formulas and always get confused in choosing which one to use. so i alwaus count, and use simple reasoning. here is how i did it.

a) how many ways there are to divide 4 people into two group of 2? that's simple counting: (ab,cd) (ac,bd) (ad,bc) - total 3. b) how many ways there are to divide 6 people into 3 groups of 2? there are 5 possibilities for member a: (ab) (ac) (ad) (ae) (af) for each such possibility we are left with 4 poeple that need to be grouped into 2 groups of 2. but we already did that (!!!!!!) so for each of the 5 options we have 3 possible completions. total of 5*3=15 options.

c) how many ways to divide 8 people into 4 teams of 2? use the same reasoning: for person a there are 7 possibilities. and for each such possibility we are eft with dividing 6 people to 3 pairs, which we laready did. hence the answer is 7*(5*3) = 105

with this reasoning (if you followed it fully....) you can answer in no time how many ways there are to divide 12 people into 6 pairs, or 20 people into 10 pairs...

Did you just create it or is it actually a formulae