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Permutation and Combination

When we talk of permutations and combinations, we often
use the two terms interchangeably. In mathematics, however, the
two have very specific meanings, and this distinction often causes
problems.

In brief, the permutation of a number of objects is the number

of different ways they can be ordered; i.e. which is first, second,
third, etc. If you wish to choose some objects from a larger number
of objects, the way you position the chosen objects is also
important. With combinations, on the other hand, one does not
consider the order in which objects were chosen or placed, just
which objects were chosen. We could summarize permutations
and combinations (very simplistically) as

Permutations - positioning is important

Combinations - choosing is important

This may help you to remember which is which. In other words,

Permutations are for lists (order matters).

Combinations are for groups (order doesn’t matter).

A formal definition is as follows:

Permutation: Permutation means arrangement of things. The word

arrangement is used, if the order of things is considered.

Combination: Combination means selection of things. The word

selection is used, when the order of things has no importance.
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Example: Suppose we have to form a number consisting of three

digits using the digits 1,2,3,4, to form this number the digits have
to be arranged. Different numbers will get formed depending
upon the order in which we arrange the digits. This is an example
of Permutation.

Now suppose that we have to make a team of 11 players out

of 20 players, this is an example of combination, because the order
of players in the team will not result in a change in the team. No
matter in which order we list out the players the team will remain
the same. For a different team to be formed at least one player
will have to be changed.
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Now let us look at two fundamental principles of counting:

Addition Rule: If an experiment can be performed in ‘n’ ways, and

another experiment can be performed in ‘m’ ways then either of
the two experiments can be performed in (m+n) ways. This rule
can be extended to any finite number of experiments.

Example: Suppose there are 3 doors in a room, 2 on one side and

1 on the other side. If a man wants to go out from the room,
obviously he has ‘3’ options for it, he can come out by door ‘A’ or
door ‘B’ or door ’C’.
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Multiplication Rule: If a work can be done in m ways, another

work can be done in ‘n’ ways, and then both of the operations can
be performed in (m * n) ways. It can be extended to any finite
number of operations.

Example: Suppose a man wants to cross-out a room, which has 2

doors on one side and 1 door on the other side. He has 2 x 1 = 2
ways for it.
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Factorial n

The product of first ‘n’ natural numbers is denoted by n!

n! = n.(n-1).(n-2)………………..3.2.1.
5! = 5 x 4 x 3 x 2 x 1 = 120

0! = 1

n! = n * (n-1)!
(n-1)! = [n * (n-1)!] / n = n! / n
Putting n = 1, we have
0! = 1! / 1
0! = 1

Factorial of a negative number is not defined.

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Permutations
Number of permutations of ‘n’ different things taken ‘r’ at a
time is given by, nPr = n! / (n-r)!

Proof: Say we have ‘n’ different things a1, a2,……, an.

Clearly the first place can be filled up in ‘n’ ways.

Number of things left after filling-up the first place = n-1.
So the second-place can be filled-up in (n-1) ways.
Now number of things left after filling-up the first and second
places = n – 2.
Now the third place can be filled-up in (n-2) ways.

Thus number of ways of filling-up first-place = n

Number of ways of filling-up second-place = n-1
Number of ways of filling-up third-place = n-2
Number of ways of filling-up rth place = n – (r-1) = n-r+1

By Multiplication – Rule of counting, total number of ways of

filling up, first, second and so on till rth place together is n (n-1) (n-
2) ------------ (n-r+1).
Hence:

nPr = n (n-1)(n-2) --------------(n-r+1)

= [n(n-1)(n-2)----------(n-r+1)] [(n-r)(n-r-1)-----3.2.1.] / [(n-
r)(n-r-1)] ----3.2.1]
nPr = n!/(n-r)!
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Question: How many different signals can be made by 5 flags from

8 flags of different colors?

Answer: Number of ways of arranging 5 flags out of 8 flags

= 8P5
= 8! / (8-5)!
= 8 x 7 x 6 x 5 x 4 = 6720.
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Number of permutations of ‘n’ different things taken all at a

time is given by nPn = n!

Proof: Now we have ‘n’ objects and n places.

Number of ways of filling-up first-place = n

Number of ways of filling-up second-place = n-1
Number of ways of filling-up third-place = n-2
Number of ways of filling-up rth place, i.e. last place =1

Number of ways of filling-up first, second, so on till nth place is = n

(n-1) (n-2) ------ 2.1.
nPn = n!

Question: How many words can be made by using the letters of

the word “SIMPLETON” taken all at a time?

Answer: There are ‘9’ different letters of the word “SIMPLETON”.

Number of Permutations taking all the letters at a time = 9P9 = 9! =
362880.
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Number of permutations of n things, taken all at a time, in

which ‘p’ are of one type, ‘q’ of them are of second type, ‘r’ of
them are of third type, and the rest are all different is given by
n! / (p! * q! * r!).

Question: In how many ways can the letters of the word “Pre-
University” be arranged?

Answer: In the word Pre-University, there are 13 letters.

The total numbers of permutations possible is 13!
Out of the 13 letters, the letters ‘I’, ‘e’ & ‘r’ are repeated
twice.
Hence, the total number of permutations is

= 13! / (2! * 2! * 2!)

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Number of permutations of n things, taken ‘r’ at a time when

each thing can be repeated r times is given by nr.

Proof: Number of ways of filling up the first place is = n

Since repetition is allowed, so the,
Number of ways of filling up the second place is = n
Number of ways of filling up the third place is = n
Number of ways of filling up the rth place is = n
Hence total number of ways in which first, second so on till
the rth place can be filled-up in
= n x n x n ------------- r factors.
= nr

Question: A child has 3 pockets and 4 coins. In how many ways

can he put the coins in his pocket?

Answer: First coin can be put in 3 ways, similarly second, third

and fourth coins also can be put in 3 ways. So total number of
ways = 3 x 3 x 3 x 3 = 34 = 81.
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Restricted – Permutations
(a) Number of permutations of ‘n’ things, taken ‘r’ at a time,
when a particular thing is to be always included in each
arrangement = r * n-1Pr-1

(b) Number of permutations of ‘n’ things, taken ‘r’ at a time,

when a particular thing is fixed = n-1Pr-1

(c) Number of permutations of ‘n’ things, taken ‘r’ at a time,

when a particular thing is never taken = n-1Pr

(d) Number of permutations of ‘n’ things, taken ‘r’ at a time,

when ‘m’ specified things always come together = m! * (n-m+1)!

(e) Number of permutations of ‘n’ things, taken all at a time,

when ‘m’ specified things always come together = n! - [m! * (n-
m+1)!]

Example: How many words can be formed with the letters of the
word ‘OMEGA’ when,

(a) ‘O’ and ‘A’ occupy end places

(b) ‘E’ being always in the middle
(c) Vowels occupying odd-places
(d) All Vowels being never together.
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Answer:

(a) When ‘O’ and ‘A’ occupy end places, this implies M.E.G.(OA)
Here (OA) are fixed,
Hence M, E, G can be arranged in 3! ways,
But (O, A) can be arranged themselves is 2! ways.
=> Total number of words
= 3! * 2!
= 12

(b) When ‘E’ is fixed in the middle, this implies O.M.(E).G.A.

Hence the four letters O.M.G.A. can be arranged in 4! = 24 ways.

(c) Three vowels (O, E, A) can be arranged in the odd-places (1st,

3rd and 5th) = 3! Ways.
And two consonants (M, G) can be arranged in the even-place
(2nd, 4th) = 2! Ways.
=> Total number of ways = 3! * 2! = 12 ways.

(d) Total number of words = 5! = 120

If all the vowels come together, then we have: (O.E.A.), M, and G.
These can be arranged in 3! ways.
But (O, E, A) can be arranged themselves in 3! ways.
=> Number of ways, when vowels come-together
= 3! * 3!
= 36 ways
=> Number of ways, when vowels being never-together
= 120 – 36
= 84 ways.
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Combination
Number of Combination of ‘n’ different things, taken ‘r’ at a
time is given by nCr= n! / r! * (n-r)!

Proof: Each combination consists of ‘r’ different things, which can

be arranged among them in r! ways.
For one combination of ‘r’ different things, number of
arrangements = r!
For nCr combination, number of arrangements = r! * nCr
Total number of permutations = r! * nCr
But number of permutations of ‘n’ different things, taken ‘r’
at a time = nPr

From the above two statements we have,

nPr = r! * nCr
n! / (n-r)! = r! * nCr
nCr = n! / r! * (n-r)!

nCr = nCn-r

nCr = n! / r! * (n-r)!
nCn-r = n! / (n-r)! * (n-(n-r))!
= n! / (n-r)! * r!
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Restricted – Combinations
(a) Number of combinations of ‘n’ different things taken ‘r’ at a
time, when ‘p’ particular things are always included = n-pCr-p

(b) Number of combination of ‘n’ different things, taken ‘r’ at a

time, when ‘p’ particular things are always to be excluded = n-
pCr

Example: In how many ways can a cricket-eleven be chosen out of

15 players, if
(i) A particular player is always chosen.
(ii) A particular player is never chosen.

Answer:

(i) A particular player is always chosen, it means that 10 players

are selected out of the remaining 14 players.
Required number of ways = 14C10
= 14! / (4! * 10!)
= 1365

(ii) A particular player is never chosen, it means that 11 players

are selected out of 14 players.
Required number of ways = 14C11
= 14! / (11! * 3!)
= 364
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Number of ways of selecting one or more things from ‘n’

different things is given by 2n-1

Proof: Number of ways of selecting 1 thing, out of n things = nC1

Number of ways of selecting 2 things, out of n things = nC2
Number of ways of selecting 3 things, out of n things = nC3
Number of ways of selecting n things out of n things = nCn

Total number of ways of selecting one or more things out of n

different things

= nC1 + nC2 + nC3 + ------------- + nCn

= nC0 + nC1 + nC3 + ------------- + nCn – nC0
= 2n – 1

Example: John has 8 friends. In how many ways can he invite one
or more of them for dinner?

Answer: John can select one or more than one of his 8 friends.
Required number of ways = 28 – 1
= 255
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Number of ways of selecting zero or more things from ‘n’

identical things is given by (n+1).

Example: In how many ways, can zero or more letters be selected

from the letters AAAAA?

Answer: Number of ways of,

Selecting zero 'A's = 1

Selecting one 'A's = 1
Selecting two 'A's = 1
Selecting three 'A's = 1
Selecting four 'A's = 1
Selecting five 'A's = 1

Required number of ways = 6

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Number of ways of selecting one or more things from ‘p’

identical things of one type ‘q’ identical things of another type,
‘r’ identical things of the third type and ‘n’ different things is
given by [(p+1) (q+1) (r+1)2n] – 1

Example: Find the number of different choices that can be made

from 3 apples, 4 bananas and 5 mangoes, if at least one fruit is to
be chosen.

Answer:

Number of ways of selecting apples = (3+1) = 4 ways.

Number of ways of selecting bananas = (4+1) = 5 ways.
Number of ways of selecting mangoes = (5+1) = 6 ways.

Total number of ways of selecting fruits = 4 x 5 x 6

But this includes, when no fruits i.e. zero fruits is selected
Number of ways of selecting at least one fruit = (4x5x6) -1 = 119.

Note: There was no fruit of a different type, hence here n=0,

2n = 20 = 1.
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Number of ways of selecting ‘r’ things from ‘n’ identical things

is ‘1’.

Example: In how many ways 5 balls can be selected from ‘12’

identical red balls?

Answer: The balls are identical, total number of ways of selecting

5 balls = 1.

Example: How many numbers of four digits can be formed with

digits 1, 2, 3, 4 and 5?

Answer: Here n = 5 and r = 4

Required number is 5P4
= 5! / 1!
=5x4x3x2x1
= 120
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Exercise
1) Find the number of permutations and combinations which can
be made by taking 4 items at a time from 6 given distinct items,
without repetition?

2) In how many ways can 3 persons be seated in 5 chairs?

3) From 10 persons waiting in a queue, in how many ways can a

selection of 6 be made so that

(i) A specified person is always included?

(ii) A specified person is always excluded?

4) Fairplay Athletics club has 6 coaches, viz, A, B, C, D, E and F. A

panel of coaches comprising of 3 members has to be formed.
(a) In how many ways can the panel be formed?
(b) How many of these panels always include coach E?
(c) In how many of these panels will coach C be excluded?
(d) In how many ways can the panel be formed if coach C & coach
F should be there together if at all any one of them is there?
(e) In how many ways can the panel be formed if coach E & coach
B cannot be together on the panel?
(f) In how many ways can the panel be formed if exactly one
among A & B should be included?
(g) In how many ways can the panel be formed if it is known that
coach D will not be on the panel if coach A is there on it?

5) How many different permutations can be made out of the

letters of the word, ‘ASSISTANTS’ taken all together?
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6) All the odd numbers from 1 to 9 are written in every possible

order. How many numbers can be formed if repetition is not
allowed?

7) In how many ways can 2 cards be drawn from a full pack of 52

cards such that both the cards are red?

8) After group discussion and interview 6 candidates were

selected for admission in a college. But unfortunately the number
of seats left is 2. So it was left to the principal to select 2
candidates out of them. In how many ways can he select 2
candidates?

9) Ram buys 7 novels from a book fair. Shyam buys 8 novels from
the fair, none of which is common with those bought by Ram. They
decide to exchange their books one for one. In how many ways can
they exchange their books for the first time?

10) How many different words can be made from the word
‘education’ so that all the vowels are always together?

11) In a cultural festival, 6 programmes are to be staged, 3 on a

day for 2 days. In how many ways could the programmes be
arranged?

12) How many committees of 5 members each can be formed from

8 official and 4 non-official members such that each committee
consists of 3 official and 2 non-official members?

13) In an entrance test, a candidate is required to attempt a total

of 4 questions which are to be attempted from 2 sections each
containing 5 questions. The maximum number of questions that
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he can attempt from any section is 3. In how many ways can he

answer in the test?

14) A palindrome is a number that reads the same left to right as it

does from right to left, such as 131. How many 6 digit palindromes
are there which are even?

15) Five distinct pairs of shoes are displayed. In how many ways
can three shoes be selected containing a matching pair?

16) How many 4 digit numbers can be made with the digits 0, 1, 2
& 7 so that at least one of the digits is repeated in every number?

17) The number of ways in which 5 boys and 5 girls can form a
circle such that the boys and girls alternate is?