How many different committees each composed of 2 Republicans : GMAT Problem Solving (PS)
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# How many different committees each composed of 2 Republicans

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How many different committees each composed of 2 Republicans [#permalink]

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11 Nov 2009, 07:33
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How many different committees each composed of 2 Republicans and 3 Democrats can be formed from a group of 4 Republicans and 5 Democrats?

A) 12
B) 24
C) 60
D) 72
E) 120

thanxx
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Feb 2012, 11:39, edited 1 time in total.
Edited the question and added the OA
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11 Nov 2009, 07:46
How many different committees each composed of 2 Republicans and 3 Democrats can be formed from a group of 4 Republicans and 5 Democrats?
A) 12
B) 24
C) 60
D) 72
E) 120

thanxx

4C2 (# of selections of 2 Republicans out of 4) times 5C3 (# of selections of 3 Democrats out of 5)=4C2*5C3=6*10=60

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11 Nov 2009, 07:48
How many different committees each composed of 2 Republicans and 3 Democrats can be formed from a group of 4 Republicans and 5 Democrats?
A) 12
B) 24
C) 60
D) 72
E) 120

thanxx

c. 60

Let's take the Republicans. How many combinations of 2 can be created.
There are four total so: 4!/2!2! = 6

Democrats: 5!/3!2! = 10

6 x 10 = 60
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11 Nov 2009, 07:49
sorry, still dont get it .. whats 4C2 .. 5C3 .. how do u calculate that?
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11 Nov 2009, 07:51
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c. 60

Let's take the Republicans. How many combinations of 2 can be created.
There are four total so: 4!/2!2! = 6

Democrats: 5!/3!2! = 10

6 x 10 = 60[/quote]

why are we dividing by additional 2! .. i got as far as 4!/2! and 5!/3! .. dont get why there is also the 2!
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11 Nov 2009, 08:01
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c. 60

Let's take the Republicans. How many combinations of 2 can be created.
There are four total so: 4!/2!2! = 6

Democrats: 5!/3!2! = 10

6 x 10 = 60

why are we dividing by additional 2! .. i got as far as 4!/2! and 5!/3! .. dont get why there is also the 2!

Okay, the non-mathematical way I use to remember is:

total number from the pool available (4!) / total number of people we need (2!)*total number of people we don't need(2!)

For democrats: 5 from the pool, 3 we need, 5-3 we don't need

completely non-mathematical approach but it works every time

so that extra 2! is taking away what we don't need.

try it on other problems..works for me
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11 Nov 2009, 08:05
bunuel can explain the mathematical approach really well i'm sure
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11 Nov 2009, 08:06
thank you, i'll remember that
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11 Nov 2009, 08:10
Since the question mentions 'different' committees, we need to use Combinations

Therefore, as others have rightly pointed out, 4C2 * 5C3 (we are 'anding' here)

Hence C
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11 Nov 2009, 09:11
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sorry, still dont get it .. whats 4C2 .. 5C3 .. how do u calculate that?

To deal with such question you should be aware of the combinatorics or fundamental principles of counting. So below is the short review:

FUNDAMENTAL PRINCIPLE OF COUNTING

Principle of Multiplication
If an operation can be performed in ‘m’ ways and when it has been performed in any of these ways, a second operation that can be performed in ‘n’ ways then these two operations can be performed one after the other in ‘mxn’ ways.

If an operation can be performed in ‘m’ different ways and another operation in ‘n’ different ways then either of these two operations can be performed in ‘m+n’ ways.(provided only one has to be done) FACTORIAL ‘n’: The product of the first ‘n’ natural numbers is called factorial n and is deonoted by n! i.e, n!=1×2x3x…..x(n-1)xn.

PERMUTATION
Each of the arrangements which can be made by taking some or all of a number of items is called a Permutation.
The permutation of three items a, b,c taken two at a time is ab, ba, bc, cb, ac, ca. Since the order in which items are taken is important, ab and ba are counted as two different permutations. The words ‘Permutation’ and ‘Arrangement’ are synonymous and can be used interchangeably
A number of permutation of n things taking r at a time(where n ≤ r) is denoted by nPr and read as nPr.
$$nPr = \frac{n!}{(n-r)!}$$

COMBINATION
Each of the group or selection which can be made by taking some or all of a number of items is called a Combination.
In combination, the order in which the items are taken is not considered as long as the specific things are included. The combination of three items a, b and c taken two at a time are ab, bc and ca. Here ab and ba are not considered separately because the order in which a and b are taken is not important but it is only required that combination including a and b is what is to be counted. The words ‘Combination’ and ‘Selection’ are synonymous.
The number of combinations of n dissimilar things taken r at a time is denoted by nCr and read as nCr.
$$nCr = \frac{n!}{((n-r)!*r!)}$$.

IMPORTANT RESULTS FOR COMBINATION UNDER DIFFERENT CONDITION
1. $$nCr =\frac{n!}{r!(n-r)!}$$

2.$$nCr = nCn-r$$

3. $$nCr + nCr-1 = n+1Cr$$

4. If $$nCr = nCs$$ => $$r = s$$ or $$n = r + s$$

DISTRIBUTION OF THINGS INTO GROUPS
1. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is $$\frac{(mn)!}{(n!)^m*m!}$$.

2. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important is $$\frac{(mn)!}{(n!)^m}$$

3. The number of ways in which (m+n+p) things can be divided into three different groups of m,n, and p things respectively is $$\frac{(m+n+p)!}{m!n!p!}$$

4. The required number of ways of dividing 3n things into three groups of n each $$\frac{(3n)!}{3!*(n!)^3}$$

When the order of groups has importance then the required number of ways = $$\frac{(3n)!}{(n!)^3}$$.

DIVISION OF IDENTICAL OBJECTS INTO GROUPS
The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0, 1, 2 or more items $$n+r-1Cr-1$$

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is $$n-1Cr-1$$

RESULTS TO REMEBER
In a plane if there are n points of which no three are collinear, then
1. The number of straight lines that can be formed by joining them is $$nC2$$.

2. The number of triangles that can be formed by joining them is $$nC3$$.

3. The number of polygons with k sides that can be formed by joining them is $$nCk$$.

You can also check excellent topic by Walker, theory, and links to combinations questions at: combinations-permutations-and-probability-references-56486.html
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11 Nov 2009, 10:09
WOW! ... darn it i was hoping i didnt have to know that much for the gmat ...
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11 Nov 2009, 10:18
WOW! ... darn it i was hoping i didnt have to know that much for the gmat ...

Don't worry it just seems complicated but when you start to practice in combination problem, it'll become easier after some time. You should see Walker's topic and go through the questions there from easy to hard. Any question you have, post and people from the forum will always be glad to help.

Cheer up!
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11 Nov 2009, 10:56
Bunuel, you're a life saver. Thanks!

But just one little question, are these points to remember and results to remember important?[/b] I mean have u seen questions in GMATprep or in the actual exam that need these relations? ( permutation and combination with with triangles, lines, etc)
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11 Nov 2009, 11:29
zaarathelab wrote:
Bunuel, you're a life saver. Thanks!

But just one little question, are these points to remember and results to remember important?[/b] I mean have u seen questions in GMATprep or in the actual exam that need these relations? ( permutation and combination with with triangles, lines, etc)

Well not everything is needed for GMAT. But I've seen the questions claimed to be real GMAT type which used some of the staff you've mentioned:

pentagon-problem-86284.html?highlight=pentagon
http://gmatclub.com:8080/forum/viewtopi ... w=previous

OR
There are 25 points on a plane which 7 are collinear. How many quadrilaterals can be formed from these points?

There are 12 points in a plane out of which 4 are collinear. How many straight lines can be formed by joining these points in pairs?

Out of 18 points in a plane, no 3 are in the same straight line except 5 points which are collinear. How many triangles can be formed by joining them?

OR DS:
S is a set of points in the plane. How many distinct triangles can be drawn that have three of the points in S as vertices?
(1) The number of distinct points in S is 5.
(2) No three of the points in S are collinear.

How many triangles can be formed using 8 points in the given plane?
(1) A triangle is formed by joining 3 distinct points in the plane
(2) Out of the 8 given points 3 are collinear.

I must say that theses questions claimed to be real GMAT, but I don't know for sure.
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Re: How many different committees each composed of 2 Republicans [#permalink]

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31 May 2015, 03:39
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Re: How many different committees each composed of 2 Republicans [#permalink]

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16 Sep 2016, 09:32
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Re: How many different committees each composed of 2 Republicans   [#permalink] 16 Sep 2016, 09:32
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