|An Excerpt from the 'Winners' Guide to GMAT Math- Part II
The factorial of a number is the product of all the positive integers from 1 upto the number. The factorial of a given integer n is
usually written as n! and n! denotes the product of the first n natural number. -
n! = n x (n – 1) x (n – 2) x ……… x 1
n! = n (n – 1)
0! = 1 as a rule.
Note : Factorial is not defined for improper fractions or negative integers.
then such an arrangement is called a permutation of n objects taken r at, a time.
Permutations is denoted by nPr or (n, r)
e.g., If it is required to seat 5 men and 4 women in a row such that women occupy the even places, in how many ways can this be
In a row of 9 positions, there are four places, and exactly 4 women to occupy them, which is possible in 4! ways. The remaining S
places can be filled up by 5 men in 5! ways, Total number of seating arrangements = 4! 5! = 24 x 120 = 2880
Important Permutation Rules:
(i) The total number of arrangements of n things taken r at a time in which a particular thing always occurs.
e.g., The number of ways in which 3 paintings can be arranged in an exhibition from a set of five, such that one is always included.
number of ways 3. 5-1P3-1 = 3.4P2 = 36
or 3! (4C2) = 6.6 = 36
(ii) The total number of permutations of n distinct things taken r at a time in which a particular thing never occurs = n-1Pr
e.g., The number of ways in which 3 paintings from a set of five, can be displayed for a photo-shoot, such that one painting is never
= 5-1P3 = 4P3 ways = 24
It can be observed that
rn-1Pr-1 + n-1Pr = nPr
(iii) The number of permutations of n different objects taken r at a time, when repetitions are allowed, is nr. The f place can be filled
by any one of the n objects in ‘n’ ways. Since repetition is allowed the second place can be filled in ‘n’ ways again. Thus, there are n
x n x n r times ways = nr ways to fill first r positions.
Suppose four numbers 1, 2, 3, 4 are to be arranged in the form of a circle.
The arrangement is read in anticlockwise direction, starting from any point as 1432, 4321, 3214 or 2143.
These four usual permutation correspond to one circular permutation.
Thus circular permutations are different only when the relative order of objects to be arranged is changed.
Each circular permutation of n objects corresponds to n Linear permutations depending on where (of the n positions) we start.
This can also be though of as keeping the position of one out of n objects fixed and arranging remaining n – 1 in (n – 1)! ways.
If r objects are to be chosen from n, where r ≤ n and the order of selecting the r objects is not important then such a selection is
called a combination of n objects taken r at a time and denoted by
In a permutation the ordering of objects is important while in a combination it is immaterial. e.g.,
AB and BA are 2 different Permutations but are the same combination.
Usually (except in trivial cases) the number of permutations exceeds the number of combinations. Trivial cases are when r = 0 or 1.
e.g., If there are 10 persons in a party, and if every two of them shake hands with each other, how many handshakes happen in the
SoIn: When two persons shake hands it is counted as 1 handshake and not two hence here we have to consider only combinations.
2 people can be selected from 10 in 10C2 ways.
Hence, number of handshake =
1. nCr = nCn – r
2. nCo = nCn = 1
3. n+1Cr = nCr + nCr – 1
4. n+1Cr+1 = nCr+1 + n–1Cr + n–1Cr-1
5. nPr = r! nC
6. The total no. of combinations of ‘n’ things taken some or all at a time nc = nC1 + nC2 + ……nCn = 2n – 1
Important Combination Rules
1. The number of combinations of ‘n’ things taken ‘r’ at a time in which p particular thin will always occur = n-pCr-p P things are
definitely selected in 1 way. The remaining r – p things can be selected from n – p things in n-pCr-p ways.
In how many ways can 7 letters be selected from the alphabet such that the vowels are always selected.
Soln : There are 5 vowels a, e, i, o, u which are selected in 1 way then possible number of ways = 26-5C7-5 = 21C2
2.The number of combinations of ‘n’ things taken ‘r’ at a time in which ‘p’ particular things never occur is n-pCr (n – p ≥ r)
p things are never to be selected.
Hence r things are to be selected from n - p in n–pCr ways It is clear that n - p ≥ r for this to be possible.
e.g. In how many ways can 7 letters be selected from the alphabet such that the vowels are never selected.
Soln : As vowels (a, e, i, o, u) are never selected. The 7 letters can be selected from (20 – 5), letters in = 26–5C7 = 21C7
3. The number of ways of dividing (partitioning) •n distinct things into r distinct groups, such that some groups can remain empty = rn
One object ran be put into r partitions in r ways
\ objects can be partitioned in r x r x r .... n times = rn ways
i) In how many ways can 11 identical white balls and 9 black bells be arranged in a row so that no two black balk are together?
The 11 white balls can be arranged in 1 way (all are identical)
The 9 black balls can be arranged in the 12 places in 12P9 ways
ii) In how many ways can they be arranged if black balls were identical? (all other conditions remaining same)
The 11 white balls can be arranged in 1 way.
The 9 black balls can be arranged in the 12 places in 12C9 ways.
Thus number of arrangements = 12C9
iii) In how many ways can they be arranged if all the balls are different. (all other conditions remaining same)
The 11 white balls can be arranged in 11! ways
The 9 black balls can be arranged in the 12 places in 12! ways.
Total number of arrangements = 11!12!/3!
In a multiple choice test there are 50 questions each having 4 options, which are equally likely.
In how many ways can a student attempt the questions in the test?
Each question can be attempted in 4 ways and not attempted in 1 way. \Each question can be attempted or unattempted in 5 ways.
Thus 50 questions can be attempted or attempted in 550 ways.
This will include the case when no questions are attempted.
\The student can attempt the paper in (5 to the power of 50) – 1 ways.
How many 4 digit numbers can be formed from the, digits 1, 5, 2, 4, 2, 9, 0, 4, 2
i) with repetition of digits.
ii) without repetition of digits.
i) In the given set 4 is repeated twice and 2 thrice
\Number of distinct digits = 6
The 4 digit number can be formed in 5.63 ways when repetition is allowed.
Position I can be filled in 5 ways, (as it cannot have O)
The remaining 3 positions can be filled in 6 ways each.
Hence number of numbers = 5.63 = 1080
ii). Position I can be filled in 5 ways.
Position II can be filled in 5 ways (it can contain any of 5 digits except the one in position 1
Thus number of such numbers = 5 x 5 x 4 x 3 = 300
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