There are two cases of circular-permutations:-

(a) If clockwise and anti clock-wise orders are different, then total number of circular-permutations is given by (n-1)!

(b) If clock-wise and anti-clock-wise orders are taken as not different, then total number of circular-permutations is given by (n-1)!/2!

**Proof**(a):

(a) Let’s consider that 4 persons A,B,C, and D are sitting around a round table

Shifting A, B, C, D, one position in anticlock-wise direction, we get the following agreements:-

Thus, we use that if 4 persons are sitting at a round table, then they can be shifted four times, but these four arrangements will be the same, because the sequence of A, B, C, D, is same. But if A, B, C, D, are sitting in a row, and they are shifted, then the four linear-arrangement will be different.

Hence if we have ‘4’ things, then for each circular-arrangement number of linear-arrangements =4

Similarly, if we have ‘n’ things, then for each circular – agreement, number of linear – arrangement = n.

Let the total circular arrangement = p

Total number of linear–arrangements = n.p

Total number of linear–arrangements

= n. (number of circular-arrangements)

Or Number of circular-arrangements = 1 (number of linear arrangements)

n = 1( n!)/n

circular permutation = (n-1)!

**Proof**
(b) When clock-wise and anti-clock wise
arrangements are not different, then observation can be
made from both sides, and this will be the same. Here two
permutations will be counted as one. So total
permutations will be half, hence in this case.

Circular–permutations = (n-1)!/2

**Note**:
Number of circular-permutations of ‘n’
different things taken ‘r’ at a time:-

(a) If clock-wise
and anti-clockwise orders are taken as different, then
total number of circular-permutations =
^{n}P_{r }/r

(b) If clock-wise and
anti-clockwise orders are taken as not different, then
total number of circular – permutation =
^{n}P_{r}/2r

Example: How many necklace of 12 beads each can be made from 18 beads of different colours?

Ans. Here clock-wise and anti-clockwise arrangement s are same.

Hence total number of
circular–permutations:
^{18}P_{12}/2×12

= 18!/(6 x 24)

__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-1} P_{r-1}

(b) Number of
permutations of ‘n’ things, taken ‘r’
at a time, when a particular thing is fixed: = ^{n-1}
P_{r-1}

(c) Number of
permutations of ‘n’ things, taken ‘r’
at a time, when a particular thing is never taken: = ^{n-1}
P_{r.}

(d) Number of permutations of ‘n’ things, taken ‘r’ at a time, when ‘m’ specified things always come together = m! x ( n-m+1) !

(e) Number of permutations of ‘n’ things, taken all at a time, when ‘m’ specified things always come together = n ! – [ m! x (n-m+1)! ]

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

(i) ‘O’ and ‘A’ occupying end places.

(ii) ‘E’ being always in the middle

(iii) Vowels occupying odd-places

(iv) Vowels being never together.

Ans.

(i) When ‘O’ and ‘A’ occupying end-places

=> 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! x 2! = 12 ways.

(ii) When ‘E’ is fixed in the middle

=> O.M.(E), G.A.

Hence four-letter O.M.G.A. can be arranged in 4! i.e 24 ways.

(iii) Three
vowels (O,E,A,) can be arranged in the odd-places (1^{st},
3^{rd} and 5^{th})
= 3! ways.

And two consonants (M,G,)
can be arranged in the
even-place
(2^{nd}, 4^{th}) = 2
! ways

=> Total number of ways= 3! x 2! = 12 ways.

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

If all the vowels come together, then we have: (O.E.A.), M,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! x 3!

= 36 ways

=> Number of ways, when vowels being never-together

= 120-36 = 84 ways.

Number of Combination of ‘n’ different things, taken ‘r’ at a time is given by:-

^{n}C_{r}= n!
/ r ! x
(n-r)!

Proof: Each combination consists of ‘r’ different things, which can be arranged among themselves in r! ways.

=> For one combination of ‘r’ different things, number of arrangements = r!

For ^{n}C_{r}
combination number of
arrangements:
r ^{n}C_{r}

=> Total number of
permutations = r! ^{n}C_{r}
—————(1)

But number of permutation of ‘n’ different things, taken ‘r’ at a time

= ^{n}P_{r}
——-(2)

From (1) and (2) :

^{n}P_{r}
= r! . ^{n}C_{r}

or
n!/(n-r)! = r! . ^{n}C_{r
}

or ^{n}C_{r}
= n!/r!x(n-r)!

**Note: **^{n}C_{r}
= ^{n}C_{n-r}

or ^{n}C_{r}
= n!/r!x(n-r)! and ^{n}C_{n-r} = n!/(n-r)!x(n-(n-r))!

= n!/(n-r)!xr!

__Restricted –
Combinations__

(a) Number of
combinations of ‘n’ different things taken
‘r’ at a time, when ‘p’ particular
things are always included = ^{n-p}C_{r-p}.

(b) Number of
combination of ‘n’ different things, taken
‘r’ at a time, when ‘p’ particular
things are always to be excluded = ^{n-p}C_{r}

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 is never chosen.

Ans:

(i) A particular player is always chosen, it means that 10 players are selected out of the remaining 14 players.

=. Required number of ways
= ^{14}C_{10} = ^{14}C_{4}

= 14!/4!x19! = 1365

(ii) A particular players is never chosen, it means that 11 players are selected out of 14 players.

=> Required number
of ways = ^{14}C_{11}

= 14!/11!x3! = 364

(iii) Number of ways of
selecting zero or more things from ‘n’
different things is given by:- 2^{n}-1

**Proof**: Number of ways of selecting
one thing, out of n-things = ^{n}C_{1}

Number of selecting two
things, out of n-things =^{n}C_{2}

Number of ways of selecting
three things, out of n-things =^{n}C_{3}

Number of ways of selecting
‘n’ things out of ‘n’ things = ^{n}C_{n}

=>Total number of ways of selecting one or more things out of n different things

= ^{n}C_{1}
+ ^{n}C_{2} + ^{n}C_{3} +
————- + ^{n}C_{n}

= (^{n}C_{0}
+ ^{n}C_{1} + —————–^{n}C_{n})
– ^{n}C_{0}

= 2^{n} –
1
[ ^{n}C_{0}=1]

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

Ans. John can select one or more than one of his 8 friends.

=> Required number of
ways = 2^{8} – 1= 255.

(iv) 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 form the letters AAAAA?

Ans. 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 [5+1]

(V) 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)2^{n}
– 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.

Ans:

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=o

=> 2^{n
}= 2^{0}=1

(VI) 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?

Ans. 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?

Ans. Here n = 5 [Number of digits]

And r = 4 [ Number of places to be filled-up]

Required number is ^{5}P_{4}
= 5!/1! = 5 x 4 x 3 x 2 x 1