Posted on - 22-02-2017

IIT JEE

Selection from distinct objects:

The number of selections from n different objects, taken at least one

= ^{n}C_{1}
+^{ n}C_{2 }+^{ n}C_{3 }+ …. +^{ n}C_{n}^{
}= 2^{n} – 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 = 2^{n}
. .

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 = 2^{n }– 1.

(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 a_{1}+a_{2}+a_{3}+….
+a_{n} objects, where a_{1 }are alike (of one kind ), a_{2}
are alike (of second kind ) and so on …. a_{n} are alike (of nth kind
), is [(a_{1}+1)(a_{2}+1)(a_{3}+1) ….. (a_{n}+1)]
– 1.

Selection when both identical and distinct objects are present:

The number of
selections taking at least one out of a_{1}+a_{2}+a_{3}+….+a_{n}+k
objects, where a_{1} are alike (of one kind), a_{2} are alike
(of second kind) and so on ….. a_{n} are alike (of nth kind), and k
are distinct = [(a_{1}+1)(a_{2}+1)(a_{3}+1) …. (a_{n}+1)]
2^{k} – 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 .

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.

To find number of divisors of a given natural number greater than 1 we can write n as n =

where p_{1}, p_{2}, ... , p_{n}
are distinct prime numbers and a_{1}, a_{2},...a_{n} are positive
integers. Now any divisor of n will be of the form d = (
where 0 £ b_{i} £ a_{i} , b_{i} ∈ I , " i = 1, 2, 3, …. , n)

Here number of divisors will be equal to
numbers of ways in which we can choose b_{i}’s which can be done in (a_{1} + 1)(a_{2} + 1)...(a_{n} + 1) ways.

e.g. Let n = 360

&⇒ n = 2^{3}.3^{2}.5.

&⇒
No. of divisors of 360 = (3 + 1) (2 + 1) (1 + 1) = 24.

As in above case, sum of all the divisors =

The number of factors of a given natural number ‘n’ will be odd if and only if ‘n’ is a perfect square.

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

n = 10800 = 2^{4}
´ 3^{3}´ 5^{2}

Any divisor of n
will be of the form 2^{a} ´
3^{b} ´ 5^{c}

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.