(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