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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
A) 12
B) 24
C) 60
D) 72
E) 120
11 Nov 2009, 10:11
Kudos
Bookmarks
elinka
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.
Principle of Addition
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
General Discussion
11 Nov 2009, 08:46
Kudos
Bookmarks
elinka
4C2 (# of selections of 2 Republicans out of 4) times 5C3 (# of selections of 3 Democrats out of 5)=4C2*5C3=6*10=60
Answer: C.
