Page 185 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes Theorem’s
Number of ways of choosing zero or more things from ‘n’ different things is specified by: 2 –1
n
n
Proof: Number of ways of choosing one thing, out of n-things = C
1
n
Number of choosing two things, out of n-things = C
2
n
Number of ways of choosing three things, out of n-things = C
3
Number of ways of choosing ‘n’ things out of ‘n’ things = C
n
n
Total number of manners of choosing one or more things out of n dissimilar things
n
n
n
n
= C + C + C + —————— + C
1 2 3 n
n
n
n
n
= ( C + C + ————————- C ) – C
0 1 n 0
n
n
= 2 – 1 [ C =1]
0
Example: John has 8 friends. In how many ways can he invite one or more of them to
dinner?
Solution: John can choose one or more than one of his 8 friends.
8
 Required number of ways = 2 – 1= 255.
Theorem:
Number of ways of choosing zero or more things from ‘n’ identical things is specified by : n+1
Example: In how many ways, can zero or more letters be chosen form the letters AAAAA?
Solution: Number of ways of:
Choosing zero ‘A’s = 1
Choosing one ‘A’s = 1
Choosing two ‘A’s = 1
Choosing three ‘A’s = 1
Choosing four ‘A’s = 1
Choosing five ‘A’s = 1
 Required number of ways = 6 [5+1]
Theorem:

Number of ways of Choosing 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 – 1
n

Example: Find the number of different options that can be prepared from 3 apples,
4 bananas and 5 mangoes, if at least one fruit is to be chosen.

Solution:
Number of ways of choosing apples = (3+1) = 4 ways.
Number of ways of choosing bananas = (4+1) = 5 ways.
Number of ways of choosing mangoes = (5+1) = 6 ways.
Total number of ways of choosing fruits = 4 x 5 x 6
But this involves, when no fruits i.e. zero fruits is chosen

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