Download pea-303-unit-2-permutaion-combination-1

Permutation And
Combination
Contents:
1. Introduction
2. Fundamental Principle of Counting
2.1 Product Rule
2.2 Addition Rule
3. Permutations Introduced
4. Special Cases of Permutations
5. Geometrical arrangements
6. Combinations
7. Grouping & Distribution
8. Example Questions
9. Practice Questions
1. Introduction:
Permutations and Combinations is one of the
most logical phenomenon of mathematics wherein there
are no formulae to mug up. Rather, it tests your ability
to understand the problem and interpret the situation
logically. It is more of application of common sense.
That is why you will see that most questions can be
solved without actually knowing the techniques of
permutations and combinations.
Great news, isn’t it?
Before proceeding further, let us quickly define
Factorial!!!
Factorial of a number or n! is the product
of n consecutive natural numbers starting from 1
to n. Factorial word is represented by ‘!’ or ‘L’.
Hence, 4! is 1x2x3x4 = 24.
Note: Factorial of zero or 0!=1
2. Fundamental Principle of Counting
2.1 Product Rule
If one operation can be done in x ways and corresponding to
each way of performing the first operation, a second operation can
be performed in y ways, then the two operations together can be
performed in xy ways.
If after two operations are performed in any one of
the xy ways, a third operation can be done in z ways, then the three
operations together can be performed in xyz ways.
• Let us take an example.
A, B, C and D are four places and a traveller has to go from A to D
via B and C.
He can go from A to B in 4 ways and
corresponding to each way he can take any one of the 2
ways to reach C. Hence A to C can be reached in 4×2=8
ways.
Corresponding to each of these 8 ways of
reaching C from A, there are 3 ways to reach D and the
traveller can choose any one of them.
Hence, A to D can be reached in 4x2x3=24 ways!
Here the different operations are mutually
inclusive. It implies that all the operations are being
done in succession. In this case we use the word ‘and’
to complete all stages of operation and the meaning of
‘and’ is multiplication.
Example:
 A tricolor flag is to be formed having three
horizontal strips of three different colors. 5 colors
are available. How many differently designed
flags can be prepared?
Solution:
First strip can be coloured in 5 ways, second
strip can be coloured in any of the remaining 4
colors, and the third strip can be coloured in any of
the remaining 3 colors.
Hence, we can get 5x4x3 = 60 differently designed
flags.
2.2 Addition Rule
If there are two operations such that they can be performed
independently in x and y ways respectively, then either of the two jobs can be
done in (x + y) ways.
Let us take the example of four places A, B, C and D taken above.
There are 4 different roads from B to A and 2 different roads from B
to C. In how many ways can a person go to A or C from B? The answer is
4+2=6 ways.
Here, the different operations are mutually exclusive. It implies
either of the operations is chosen. in this case we use the word ‘or’ between
various operations and the meaning of ‘or’ is addition.
The product rule and the addition rule signify the cases of ‘and’ & ‘or’.
3. Permutations Introduced
The arrangements of a given number of things taking some or
all of them at a time are called permutations.
For example, the permutations of three alphabets x, y, z taken two at
a time are xy, xz, yx, yz, zx, zy.
A point to be noted is that arrangement or order is very
important in permutations. Hence xy is distinctly different from yx.
If r things are taken at a time out of a total of n things,
then the total number of permutations is denoted by nPr.
nP
r
=n!/(n-r)!
Now you will ask why is this so. Let’s clear this.
First object can be selected in ‘n’ ways. Second
object can be selected in (n-1) ways. Third object can be
selected in (n-2) ways. Similarly the rth object can be
selected in (n-(r-1)) = (n-r+1) ways.
Therefore the total number of ways of arranging
these ‘r’ objects
= n x (n-1) x (n-2) x (n-3) x ……x (n-r+1)
={n x (n-1) x (n-2) x ……(n-r+1) x (n-r) x (n-r-1)
x….3 x 2 x 1} / {(n-r) x (n-r-1) x…..x 1}
= n! / (n-r)!
Hence, nPr = n! / (n-r)!
Example 1:
There are 4 boxes. Find the total number of
arrangements if we can arrange only 2 boxes at a time.
Solution:
Out of 4 boxes, we are arranging 2 at a time.
So total number of arrangements possible is
4P = 4! / (4-2)! = 4! / 2! = 4x3x2x1 / 2×1 = 12
2
Let us verify. Let us name the boxes A, B, C, D.
Total number of arrangements possible are AB, BC, CD,
BA, BC, BD, CA, CB, CD, DA, DB, DC.
Permutations of n different things taken r at a time
= nPr = n! / (n-r)!
Example 2:
In the above example, what if all the 4 boxes are
selected at a time? How many arrangements are
possible then?
Solution:
Total no. of arrangements possible
= 4P4 = 4! / (4-4)! = 4! / 0! = 4! = 4x3x2x1 = 24.
Permutations of n different things taken all at a
time = nPn = n!
Example 3:
If out of the 4 boxes, one particular box should
always be selected; then how many arrangements are
possible if 3 boxes are selected at a time?
Solution:
Since one box should always be selected we have to
select 3-1 boxes out of 4-1 boxes.
This can be done in 3P2 = 3! / (3-2)! = 3! / 1! = 3x2x1 / 1
= 6 arrangements.
With each of these 6 arrangements our preselected box
can be arranged in 3×6 = 18 ways.
Wondering how?
Let us name these boxes A, B, C and D and D has
to be always present.
So now A, B and C can be arranged as AB, AC,
BA, BC, CA, CB.
With AB, D can be arranged as DAB, ADB, ABD
i.e. 3 ways.
D can be arranged with the remaining 5
arrangements similarly.
Hence in total there can be 18 arrangements.
Permutations of n different things taken r at a time,
when one particular thing always occurs = r.(n-1)P(r-1)
Example 4:
How many arrangements are possible if out of the 4 boxes – A, B,
C and D one particular box D is never selected, taken 2 at a time?
Solution:
Since D is never to be selected, we have to take into account A, B
and C.
We can arrange A, B and C taken 2 at a time in 3P2 = 3! / (3-2)! =
3! / 1! = 3x2x1 / 1 = 6 ways. i.e. AB, AC, BA, BC, CA, CB.
So when one particular item is never chosen, we just ignore it and
treat the problem as if that particular item is not present in the total
number of items.
Permutations of n different things taken r at a time
when a particular thing never occurs = (n-1)Pr.
Shortcut Tip:
We know that nPr = n! / (n-r)!
Let us say we have to find out 12P3.
12P = 12! / (12-3)! = 12! / 9!
3
= 12x11x10x…..x1⁄ 9x8x7x…1 = 12x11x10 =1320
 Instead of writing out so much, the moment you
see 12P3 you should know that you have to multiply 3
numbers.
 Starting from 12, we take in 3 numbers in the
descending order and multiply them out.
 Learn to get into the habit of writing 12P3 =12x11x10
straightaway.
 This helps in faster calculation.
4. Special Cases of Permutations:
Reap to Remember:
 With respect to fundamental principle of counting,
‘and’ stands for multiplication & ‘or’ stands for
addition
 nPr= n!/ (n-r)!
 Permutations of n different things taken r at a time
= nPr = n! / (n-r)
 Permutations of n different things taken all at a time
= nPn = n!
 Permutations of n different things taken r at a time,
when one particular thing always occurs = r.(n-1)P(r-1)
 Permutations of n different things taken r at a time
when a particular thing never occurs = (n-1)Pr
Example: 5
In in how many ways can the letters of the word WATER be arranged so
that we have a new pattern every time?’
Solution:
 This is permutation of n different things taken all at a time which is
equal to n!
Hence, total number of different arrangements possible is 5! =120.
 Another way to look at it is we have 5 places to be occupied by 5
different letters.
 The 1st place can be filled by any of the 5 letters, hence 5 ways. The
2nd place can be filled by any one of the remaining 4 letters as one
letter has already been fixed at the first place, hence 4 ways.
Similarly, the 3rd place can be filled in 3 ways and the 2nd in 2 ways.
The 5th place can be filled in only one way as there is no choice but
to fill it by the remaining 1 letter.
 So going by the product rule, this can be done in 5x4x3x2x1 = 120
ways.
Permutation of n things when some are identical or
Permutation of n things not all different:
What happens when we have to find out the number of permutations
when certain items are identical?
• If 2 exactly similar red chairs(R1 & R2) and 1 black chair(B) are
to be arranged, then please note that one cannot distinguish
between the 2 red chairs. This is to say that there is no difference
between R1 B R2 and R2 B R1 because they will both look the
same as I cannot differentiate between R1 and R2 as they are
exactly same.
• So the total no. of arrangements possible will be 3! / 2! = 3. They
will be BRR RBR RRB.
• We have divided by 2! to take care of the ‘two’ items that are
same.
If out of n things, p are exactly alike of one kind, q exactly alike of
second kind and r exactly alike of third kind and the rest are all
different, then the number of permutations of n things taken all at a
time = n! / (p!q!r!)
Example 6:
In how many ways can the letters of the word
COMMITTEE can be arranged
i. using all the letters
ii. if all the vowels are together
Solution:
i. Total letters = 9 and identical letters are 2M 2T and 2E.
So total no. of arrangements = 9! / 2!2!2!
ii. Since all vowels must appear together we consider them
as one unit. There are 4 vowels- O I E E. So now we have 5
letters. Out of these we have 2M and 2T. These 5 letters can
be arranged in 5! / 2! 2! ways.
In the group of 4 vowels, the 4 vowels can arrange
themselves in 4!/2! ways.
So total no. of words formed = 5!/2! 2! X 4!/2!
Permutations where repetitions are allowed:
While dealing with letters and digits, you will often
come across cases where repetition in permutation is
allowed or not allowed. You have to be very careful as to
what is asked for because the treatment for both the cases is
absolutely different.
Example 7:
How many numbers of 5 digits can be formed with
the digits 0,1,2,3,4
i. if the digits cannot repeat themselves
ii. if the digits can repeat themselves
Solution:
i. The 1st place (ten-thousandth place) can assume
only non-zero digits. Hence it can be occupied in 4
ways. The 2nd place can be occupied by any of the
remaining 4 digits, i.e. 4 ways. Similarly, the 3rd,
4th and 5th place in 3, 2 and 1 ways respectively.
Total no. of numbers formed = 4x4x3x2x1 = 96
ii. The 1st place (ten-thousandth place) can assume
only non-zero digits. Hence it can be occupied in 4
ways. Since repetition is allowed, the 2nd, 3rd, 4th
and 5th places can all be filled in 5 ways each i.e.
we have a choice of 5 digits (0,1,2,3,4) for each
place.
So total no. of numbers formed = 4x5x5x5x5 =
4×54 =2500
The number of permutations of n different
things taken r at a time, when each may be
repeated any number of times in each
arrangement is nr.
Example 8:
How many different four letter words can be formed (the words need not be
meaningful) using the letters of the word MEDITERRANEAN such that the first
letter is E and the last letter is R?
Solution:
MEDITERRANEAN is 13-letter word.
We have to make 4 letter words that start with an 'E' and end with 'R'.
Therefore, we have to find two more letters from the remaining 11 letters.
• Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5
letters.
• The second and third positions can either have two different letters or can have
both as same letters.
• Case 1: When the two letters are different
One has to choose two different letters from the 8 available different choices.
This can be done in 8 * 7 = 56 ways.
• Case 2: When the two letters are same
There are 3 options - the two letters can be Ns or Es or As. Therefore, 3 ways.
• Total number of possibilities = 56 + 3 = 59
Finding Rank of a Given Word:
RANK OF A WORD IN DICTIONARY!!!
 Rank of a word is the position of that word, when
we arrange the words formed by alphabets of that
given word in dictionary order. Lets see an
example.
 If we are to find the rank of the word “RANK” in
the dictionary – what does that mean?
 There are 4 alphabets in the word “RANK” so we
can form 4! = 24 words by arranging them.
 This means that if we form words by permutation
of the alphabets of the word “RANK” and form a
dictionary of these words, at what position from
the top will the word “RANK” lies.
Now to find the rank of any given word in dictionary
(Without Repeating Alphabets) –
Example – “MOTHER”
1. Arrange all the alphabets in alphabetical order like (E,
H, M, O, R, T)
2. Now in dictionary words will appear in alphabetical
order, so first words will appear starting alphabet “E“.
When E is fixed at first position, rest 5 alphabets can
be arranged in 5! = 120 ways.
3. Next starting alphabet will be “H” and again there will
be 5! = 120 words starting with “H“.
4. Now starting with “M“, and next alphabet as “E” we
will have 4!=24 words.
5. Similarly starting with “M“, and next alphabet as “H”
we will have 4!=24 words.
6. Next will be starting with “M“, and next alphabet as “O”
and next as “E” we’ll have 3!=6 words.
7. Similarly starting with “M“, and next alphabet as “O” and
next as (“H” or “R”) we’ll have 3!*2=12 words.
8. Next will be starting with “M“, and next alphabet as “O”
and next as “T” and next as “E” we’ll have 2!=4 words.
9. Next will be starting with “M“, and next alphabet as “O”
and next as “T” and next as “H” will have 2! = 4 words
but the first word will be M>O>T>H>E>R which is the
desired word.
So the rank of word MOTHER in dictionary will be 5! + 5! +
4! +4! + 3! + 3! + 3! + 2! +1 which equals 309.
So the word MOTHER will be at 309th position if we form
the whole words that can be created using the letters of
MOTHER arranged in dictionary order.
Example: INDIA (With repeating Alphabets)
Alphabetical order - ADIIN
Start with A,
A = 4! / 2! = 12 (Coz there are 2 I’s)
D = 4! / 2! = 12 (Coz there are 2 I’s)
[I]A = 3! = 6
[I]D = 3! = 6
[I]I = 3! = 6
[IN]A = 2! =2
[IND]AI = 1
[IND]IA = 1.
Summing it up gives you the rank
12 + 12 + 6 + 6 + 6 + 2 + 1 + 1 = 46.
Shortcut for finding Rank: (Without repetition)
Take the Word “SURYA”
A R S U Y // Alphabetical Order
1. 2*(4!) = 48// Search for S. Remove that word from the
list and see how many letter before S? 2 letter.
2. 2*(3!) = 12 // Search for U. Remove that letter and
count the letters before U now. Its 2.
3. 1*(2!) = 2 // Search for R. Remove that letter and
count the letters before R. Its 1.
4. 1*(1!) = 1// Search for Y. Remove that letter and count
the letter before Y. Its 1.
5. Add the whole numbers. and add 1 for last letter A.
Sum: 48 + 12+ 2 + 1 + 1 = 64.
5. Geometrical arrangements:
Circular permutation:
Sitting in a circle is not the same as sitting
in a straight line. A circle does not have any
starting point or ending point. Thus in a circular
permutation, one thing is kept fixed and the
others are then arranged relative to this fixed
item. Then it is treated like a linear arrangement.
The number of circular permutations of n
different things taken all at a time = (n-1)!
 Fix any one as reference point, the remaining other
n-1 things can be arranged in (n-1)! ways.
 What if we are taking into consideration beaded
necklace or a garland wherein clockwise and
anticlockwise arrangements are the same?
 We simply divide (n-1)! by 2 to take into account
the two same clockwise and anticlockwise
arrangements.
If the clockwise and anticlockwise orders are not
distinguishable, then the number of
permutations = (n-1)! / 2
Arrangement around a regular polygon:
If n people are to be arranged around a p sided
regular polygon, such that each side of the polygon
contains the same number of people, then the number of
arrangements possible is n!/p.
For example:
15 people are to be arranged around a pentagon
shaped table having 3 people on each side of the table,
number of arrangements will be 15!/5.
Please note if the polygon is not regular, then
the number of arrangements will be n!
irrespective of the sides of the polygon.
Special case of arrangement around a
rectangular table:
 Rectangle is a special case because though it is
not a regular polygon, it is a symmetrical
quadrilateral with opposite sides equal.
 So, if n people are to be arranged around a
rectangular table, such that there are the same
number of people on each of its 4 sides, then the
total number of arrangements possible is n!/2.
 Here 2 signifies the degree of symmetry of the
rectangle.
6. Combinations:
Difference between Permutations and
Combinations:
 Suppose there are 3 bags (A,B and C) in my home
and I want to select any 2 out of them to take with
me on my holiday. In how many ways can I make
the selection?
 Clearly I select either AB or BC or AC i.e. 3
ways.
 An important point to note is that we are talking
about selection and not order here. Obviously
whether I select AB or BA makes no difference.
Let us take one more example:
Suppose from a class of 10 students I have to
select 3 students for a play, it is a case of
Combinations. But, if I have to arrange 3
students in a line from a class of 10 students, it
is a case of Permutations.
I hope I have made it clear that in permutations
(rearrangement) order matters but in
combinations (selections) order does not matter.
We are now in a position to define Combinations!!!
Definition of Combinations:
Combinations is the selection of some or all of a
total of n number of things.
If out of n things we have to select r things
(1≤r≤n), then the number of combinations is
denoted by nCr = n!/r!(n-r)!
Combinations does not deal with the
arrangements of the selected things. This explains
division by r! which denotes the arrangement of
the selected r things.
Important relation between Permutations and
Combinations
‘r’ selected things can be arranged in r! ways.
So, r! x nCr = nPr
or, nCr = nPr / r!
or nCr = n! / r! (n-r)!
Example 9:
In a class there are 6 boys and 5 girls. In
how many ways can a committee of 2 boys and 2
girls be formed?
Solution:
2 boys can be selected out of 6 in 6C2 ways.
2 girls can be selected out of 5 in 5C2 ways.
So the selection can be made in 6C2 x 5C2 ways.
(Product Rule: ‘and’ stands for multiplication)
Example 10:
In a class there are 6 boys and 5 girls. A committee of 4 is to be
selected such that it contains at least 1 boy and 1 girl.
Solution:
There are 3 different possibilities nowi. 1 boy and 3 girls
ii. 2 boys and 2 girls
iii. 3 boys and 1 girl
In the 1st possibility, total number of combinations = 6C1 x 5C3
In the 2nd possibility, total number of combinations = 6C2 x 5C2
In the 3rd possibility, total number of combinations = 6C3 x 5C1
But only one of the above possibilities will occur; 1st OR 2nd
OR 3rd.
So the total number of required combinations
is 6C1 x 5C3 + 6C2 x 5C2 + 6C3 x 5C1
Some Important Results on Combinations:
 nCr = nCn-r
(0≤r ≤n)
 nC0 = nCn = 1
 nCr + nCr-1 = n+1Cr
If nCp = nCq , then p = q or p + q = n (p,q € W)
Restricted Combination:
 The number of combinations of ‘n’ different
things taken ‘r’ at a time subject to restriction
that p particular things i) will never occur = n-pCr
ii) will always occur = n-pCr-p
 Number of ways of selecting one or more things
from a group of n distinct things
= nC1 + nC2 + nC3 + …… + nCn = 2n – 1 .
 Number of ways of selecting zero or more things
from a group of n distinct things
= nC0 + nC1 + nC2 + nC3 + …… + nCn = 2n
7. Grouping and Distribution
This is a very important concept of
permutation and combination where some higher
order fundamentals of permutation and
combination is involved – the reason for
reserving this topic for the end.
What is the difference between grouping and
distribution?
 To distribute something, first grouping is done.
Only after you have made groups of some
objects, you might want to distribute these groups
in various places.
 For example, after you made groups of some toys,
you might want to distribute these groups among
some children.
 Or, after dividing some number of toffees into
groups, you might want to distribute these groups
into boxes.
 Just as the objects that we group can be similar or
dissimilar, so can the places that we assign these
groups to be similar or dissimilar.
While distributing groups, we need to keep
one rule in mind: We permute the groups
only if these places for distribution are
dissimilar, otherwise not.
Say, we have 2 items- X and Y and I have to
split them into two groups. There is only one
way of doing it – X goes in one group and Y in
the other. However, if I have to distribute
among 2 people A and B, then these 2 groups
can be permuted in 2! ways.
Division of dissimilar items into groups of EQUAL
SIZE
 Let’s take a very simple example. In how many
ways can you divide 4 different things (say A, B, C
and D) into two groups having two things each?
 You would like to say that we select two things out
of the four and two would be left behind, i.e. 4C2 =
6 ways. But are there really 6 ways?
Take a look. We can divide four things, A, B, C and D
into two groups of two in the following ways:
• AB – CD
• AC – BD
• AD – BC
 You can keep trying but there is no fourth way to do
it. So where have the remaining 3 ways calculated
through 4C2 = 6 disappear?
 If you look carefully, there was an overlap. When we
select 2 things out of 4, we can do it in 6 ways – AB,
AC, AD, CD, BD and BC but when we select the first
three groups, the last three get automatically selected
without having to select them separately and viceversa.
 So when we select AB, CD is automatically selected
and vice-versa.
.
This overlap will manifold itself if we increase
the number of items further
So be very careful not to apply the usual
combinations formula whenever we have to
divide into groups of equal size.
However, if the groups contain unequal
number of things, then our earlier method of
using combinations formula for selection will
be valid as will be discussed in the next
section.
 Let me now increase the objects to 5. How would you
divide these 5 distinct (dissimilar) objects into groups of
2, 2 and 1?
 The single object can be chosen in 5C1 = 5 ways. The rest
of the 4 objects can be divided into two equal groups in 3
ways as explained above. Therefore, total number of ways
= 5 x 3 = 15.
The number of ways in which mn different things
can be DIVIDED equally into m groups, each group
containing n things = (mn)!/(n!)m x 1/m!
The number of ways in which mn different things
can be DISTRIBUTED equally into m groups, each group
containing n things = (mn)! / (n!)m
Note: In the distribution, order is important hence the
divisible things can be arranged in m! ways since
things are divided into m groups.
Division of dissimilar items into groups of
UNEQUAL SIZE
 Say we have k things and we have to divide them
into 2 groups containing m and n things
respectively such that m+n =k, then this can be
done in k!/m!.n! ways.
 This is because m things can be selected out of k
things in kCm ways and when m things are taken,
n things are left to form the other group of n
things which can only be done in nCn =1 way.
 Hence the required number of ways
is kCm = m+nCm = (m+n)!/(m+n-n)!.n! = k!/m!.n!.
 We can extend the same concept for increased
number of groups as long as the number of items in
all the groups add up to the total i.e. n.
The number of ways in which n distinct things can
be DIVIDED into R unequal groups containing a1,
a2, a3, ……, aR things (different number of things
in each group and the groups are not distinct)
= nCa1 × (n-a1)Ca2 × … × (n-a1-a2-….-a(r-1))CaR
=n! / a1! a2! a3!…….aR! (here a1 + a2 + a3
+ … + aR = n)
What if there are n distinct things and we have to find out
the number of ways in which they can be distributed
among r persons such that some person get a1 things,
another person get a2 things, . . . . and similarly someone
gets aR things (each person gets different number of
things)?
Number of ways in which n distinct things can be
divided into R unequal groups containing a1, a2, a3,
……, aR things (different number of objects in each
group and the groups are distinct)
=n! / a1! a2! a3!…….aR! x R! (here a1 + a2 +
a3 + … + aR = n)
Division of IDENTICAL / SIMILAR ITEMS
into Groups
Number of ways in which n identical things can
be divided into r groups, if blank groups are
allowed i.e. each can receive zero or more
things (here groups are numbered, i.e., distinct),
where 0≤r≤n = (n+r-1)C(r-1)
Number of ways in which n identical things
can be divided into r groups, if blank groups
are not allowed i.e. each receives at least one
item (here groups are numbered, i.e., distinct),
where 1≤r≤n = (n-1)C(r-1)
Number of ways in which n identical things
can be divided into r groups so that no group
contains less than m items and more than k
(where m<k) is coefficient of xn in the
expansion of (xm + xm+1 +…..xk)r
8. Example Questions
Example 11:
Find the number of even natural numbers which have 3
digits.
(a) 450
(b) 900
(c) 500
(d) 499
Solution:
 100th place can be filled only by 1,2,3,4,5,6,7,8,9 i.e 9
ways as 0 cannot come in that place.
 10th place can be occupied by any of the digits from 0
to 9 i.e 10 ways.( as there is no bar on repeating the
digits)
 Units place can be filled up by one of 0,2,4,6,8 as the
numbers formed have to be even. i.e 5 ways.
 Thus there are 9*10*5= 450 even 3 digit numbers
Example 12:
Find the number of ways in which the letters of
the word EPIDEMIC can be arranged?
(a) 10080
(b) 0
(c) 1080 (d)958
Solution:
The word EPIDEMIC has 8 letters with 2 E's,
and 2 I's. So the possible number of words =
8!/2!.2!= 8!/4= 10080
Example 13:
7 students are to be accommodated in 5 chairs in
a row so that each chair has only one student and
the shortest student is definitely accommodated
in one chair. How many arrangements are
possible?
(a) 1600 (b) 2400 (c) 1800 (d) 2000
Solution:
Here n= 7, r= 5, So, the number of possible
arrangements = 5∗ 7−1 ! /7−5 ! = 5∗ 6 !/ 2 ! =
5∗6∗5∗4∗3∗2∗1/2∗1 = 1800
Example 14:
In how many ways can a committee of 3 men and 2 ladies
be appointed from 6 men and 4 ladies?
(a) 60
(b) 120
(c) 240
(d) 180
Solution:
Number of ways of selecting 3 men out of 6 men = 6C3
Number of ways of selecting 2 ladies out of 4 ladies = 4C2
So the number of ways of forming the committee =
6C . 4C = 20*6 = 120
3
2
Example 15:
There are 6 boys and 4 girls in a class. In how many ways can
they be seated in a row so that no 2 girls are together?
(a)720
(b) 604800 (c) 52000
(d) 820
Solution:
Let the 6 boys be 1st seated in a row with space between them
as shown.
( S- space, B- boys)
S1 B1 S2 B2 S3 B3 S4 B4 S5 B5 S6 B6 S7
The boys can be seated in 6! Ways.
As no 2 girls are to be together, they have to be seated in 7
spaces between the boys.
There are thus 7 spaces to seat the 4 girls.
This can be done in 7P4 ways.
So the required number of ways to seat the 4 girls and 6 boys
= 6! * 7P4 ways = 604800 ways
9. Practice Questions
1. In how many different ways can the letters of
the word VIKADAKAVI be arranged such
that vowels are not together?
A. 73800
B. 63800
C. 52406
D. 54000
2. What is the value of 32!/29!?
A. 29760
B. 14689
C. 25470
D. 29860
3. In how many ways can the letters of the word
INDEPENDENCE can be arranged so that the
consonants come together?
A. 6500
B. 12600
C. 18500
D. 9822
4. How many 3 digit numbers can be formed by
using the digits 3,6,9 and how many of these are
even?
A. 18,6
B. 27,9
C. 15,12
D. 20,4
5. A department had 8 male and female
employees each. A project team involving 3 male
and 3 female members needs to be chosen from
the department employees. How many different
project teams can be chosen?
A. 112896
B. 3136
C. 720
D. 112
6. In how many ways can 6 identical rings be
worn on the four fingers of the hand?
A. 6C4
B. 6C4
C. 46
D. 64
7. A Joint Politician and Police committee of 5
members is to be formed 4 Politician, 3 male
Police and 5 female Police. How many different
committees can be formed if the committee must
consist of 2 Politician, 1 male Police and 2
female Police?
A. 170
B. 152
C. 180
D. 104
8. A college has 10 basketball players. A 5member team and a captain will be selected out
of these 10 players. How many different
selections can be made?
A. 1260
B. 210
C. 10C6 × 6!
D. 10C5 × 6
9. An eight letter word is formed by using all the letters
of the word EQUATION. How many of these words
begin with a consonant and end with a vowel?
A. 3600
B. 10800
C. 2160
D. 720
10. Srinivasan and Praveen are in a horse race
with 6 contestants in total. How many different
arrangements of finishes are there if Praveen
always finishes before Srinivasan and if all the
horses finish the race?
A. 700
B. 360
C. 120
D. 24
11. A Professional courier company has three
packages to deliver to three different houses. If
the packages are delivered at random to the three
houses, how many ways are there for at least one
house to get the wrong package?
A. 3
B. 5
C. 3!
D. 5!
12. In how many ways can 6 black shoes and 6
brown shoes be arranged such that 2 particular
brown shoes are never to be together?
A. 11! × 2!
B. 9! × 90
C. 110 × 10!
D. 18 × 10!
13. In a cricket match, if Afridi scores 0, 1, 2, 3, 4
or 6 runs of a ball, then find the numbers of
different sequences in which he can score exactly
30 runs of an over. Assume that an over consists
of only 6 balls and there were no extra and no run
out.
A. 86
B. 71
C. 56
D. 65
14. A IIM-A student is required to answer 6 out
of 10 questions divided into two groups each
containing 5 questions. He is not permitted to
attempt more than 4 from each group. In how
many ways can he make the choice?
A. 210
B. 150
C. 100
D. 200
15. While packing for a business trip Mr.Jayasurya has
packed 3 pairs of shoes, 4 pants, 3 half-pants, 6 shirts, 3
sweaters and 2 jackets. The outfit is defined as consisting
of a pair of shoes, a choice of "lower wear" (either a pant
or a half-pant), a choice of "upper wear" (it could be a shirt
or a sweater or both) and finally he may or may not choose
to wear a jacket. How many different outfits are possible?
A.
B.
C.
D.
567
1821
743
1701
16. A number lock on a suitcase has 3 wheels
each labeled with 10 digits from 0 to 9. If
opening of the lock is a particular sequence of
three digits with no repeats, how many such
sequences will be possible?
A. 720
B. 760
C. 680
D. 780
17. How many lines can be drawn through 21
points on a circle?
A. 310
B. 210
C. 410
D. 570
18. A question paper has two parts, part A and
part B, each containing 10 questions. If the
student has to choose 8 from part A and 5 from
part B, in how many ways can he choose the
question?
A. 11240
B. 12240
C. 13240
D. 11340
19. How many parallelograms will be formed if
7 parallel horizontal lines intersect 6 parallel
vertical lines?
A. 42
B. 294
C. 315
D. 240
20. A polygon has 44 diagonals, then the number
of its sides are:
A. 11
B. 9
C. 7
D. 5
21. One red flag, three white flags and two blue
flags are arranged in a line such that: I. No two
adjacent flags are of the same colour and II. The
flags at the two ends of the line are of different
colours. In how many different ways can the
flags be arranged?
A. 6
B. 4
C. 10
D. 2
Ans:A
22. How many numbers are there between 100
and 1000 in which all the digits are distinct?
A. 548
B. 648
C. 748
D. 848
Ans: B
23. If there are 12 persons in a party, and if each
two of them shake hands with each other, how
many handshakes happen in the party?
A. 44
B. 66
C. 77
D. 55
Ans: B
24. If the letters of the word SACHIN are
arranged in all possible ways and these words are
written out as in dictionary, then the word
‘SACHIN’ appears at serial number:
A. 601
B. 600
C. 603
D. 602
Ans:A
25. If the letters of the word EARN are arranged
in all possible ways and these words are written
out as in dictionary, then the word ‘NEAR’
appears at serial number:
A. 7
B. 15
C. 20
D. 22
Ans: B
Thank you