All Possible Selections
Selection from distinct objects:
The number of selections from n different
objects, taken at least one
= nC1
+ nC2 + nC3 + …. + nCn
= 2n – 1.
In other words, for
every object, we have two choices i.e. either select or reject in a particular group.
Total number of choices (all possible selections) = 2.2.2 …. n times = 2n
. .
But this also
includes the case when none of them is selected and the number of such cases =
1. .
Hence the number of
selections, when at least one is selected = 2n – 1.
Selection from identical objects:
(a) The number of
selections of r objects out of n identical objects is 1.
(b) Total number of selections of zero or
more objects from n identical objects is n + 1.
(c) The total number of
selections of at least one out of a1+a2+a3+….
+an objects, where a1 are alike (of one kind ), a2
are alike (of second kind ) and so on …. an are alike (of nth kind
), is [(a1+1)(a2+1)(a3+1) ….. (an+1)]
– 1.
Selection
when both identical and distinct objects are present:
The number of
selections taking at least one out of a1+a2+a3+….+an+k
objects, where a1 are alike (of one kind), a2 are alike
(of second kind) and so on ….. an are alike (of nth kind), and k
are distinct = [(a1+1)(a2+1)(a3+1) …. (an+1)]
2k – 1.
Example.1
Let a person have 3
coins of 25 paise, 4 coins of 50 paise and 2 coins of 1 rupee. Then, in how
many ways can he give none or some coins to a beggar? Further find the number
of ways so that.
(i) he gives at
least one coin of one rupee. .
(ii) he gives at
least one coin of each kind .
Solution
Total number of ways
of giving none or some coins is (3 + 1) (4 + 1) (2 + 1) = 60 ways
(i) Number of ways
of giving at least one coin of one rupee = (3 + 1) (4 + 1) ´ 2 = 40
(ii) Number of ways
of giving at least one coin of each kind = 3 ´ 4 ´
2 = 24.
Total number of divisors
of a given natural number:
To find number of divisors
of a given natural number greater than 1 we can write n as n = 
where p1, p2, ... , pn
are distinct prime numbers and a1, a2,...an are positive
integers. Now any divisor of n will be of the form d = 
(
where 0 £ bi £ ai , bi ∈ I , " i = 1, 2, 3, …. , n)
Here number of divisors will be equal to
numbers of ways in which we can choose bi’s which can be done in (a1 + 1)(a2 + 1)...(an + 1) ways.
e.g. Let n = 360
&⇒ n = 23.32.5.
&⇒
No. of divisors of 360 = (3 + 1) (2 + 1) (1 + 1) = 24.
Sum of all the divisors of
n is given by
As in above case,
sum of all the divisors = 
Remark:
The number of factors of a given natural
number ‘n’ will be odd if and only if ‘n’ is a perfect square.
Example.1
If n = 10800, then
find the
(a)
Total
number of divisors of n
(b)
The
number of even divisors
(c)
The
number of divisors of the form 4m+2
(d)
The
number of divisors which are multiples of 15
Solution
n = 10800 = 24
´ 33´ 52
Any divisor of n
will be of the form 2a ´
3b ´ 5c
where 0 £ a £ 4, 0 £
b £ 3, 0 £ c £ 2 . For any distinct choices of a,b and c , we get a
divisor of n .
(a)
Total
number of divisors = (4 + 1) (3 + 1) (2 + 1) = 60.
(b)
For
a divisor to be even, ‘a’ should be at least one. So total number of
even divisors = 4(3+1)(2+1) = 48.
(c)
4m+2
= 2(2m+1). In any divisor of the form 4m+2, ‘a’ should be exactly 1. So
number of divisors of the form 4m + 2
= 1 ( 3+1) (2+1) = 12.
(d)
A
divisor of n will be a multiple of 15 if b is at least one and c is
at least one. So number of such divisors = ( 4+1)´3´
2 = 30.
