Question

There are 30 balls in a basket. 12 of themthere are 30 boys in a basket 12 of them are tennis Boss 8 all table tennis Boss for all cricket boss and the last all gold was what fraction of the boss are was expressed as a fraction in the simplest form​

Answer 1

Step-by-step explanation:

I. In How many ways can these scholarships be awarded?

Clearly classical scholarship can be awarded to anyone of the 3 candidates, similarly mathematical and natural science scholarship can be awarded in 5 and 4 ways respectively. So,

Number of ways of awarding three scholarshipsV= 3 X 5 X 4 = 60 -----------------------[ By Fundamental Principle of Multiplication]

II. In How many ways one of these scholarships be awarded?

Number of ways of awarding one of the three scholarships = 3 + 5 + 4 = 12------------------------[ By Fundamental Principle of Addition]

Example 2 :- A room has 6 doors. In how many ways can a person enter the room through one door and come out through a different

Number of ways coming in the room = 6

Number of ways going out of the room = 5 (He/She cannot go from the same door)

By Fundamental Principle of Multiplication--------> Coming in X Going out = 6 X 5 = 30.

Example 3 :- Five persons entered the lift cabin on the ground floor of an 8 floor house. Suppose each of them can leave the cabin independently at any floor beginning with the first. Find the total number of ways in which each of the five persons can leave the cabin

i) At any one of the 7 floors

ii) At different floors.

Let the five persons be b,c,d,e,f

I) b can leave the cabin at any of the seven floors. So he has 7 options

Similarly each of c,d,e,f also has 7 options. Thus the total number of ways in which each of the five persons can leave the cabin at any of the seven floors is

7 X 7 X 7 X 7 X7 =

II) b can leave the cabin in 7 ways. c can leave the cabin in 6 ways, since he can not leave at where b left. In the same way d has 5, e has 4, and f has 3 way.

Hence total number of ways = 7 X 6 X 5 X 4 X 3 = 2520

Example 4 :- In how many ways can 3 prizes be distributed among 4 boys, when

i) No boy gets more than one prize ?

ii) A boy may get any number of prizes ?

iii) No boy gets all the prizes ?

I) The first prize can be given away to any of the 4 boys, hence there are 4 ways to distribute first prize.

The second prize can be given away to any of the remaining 3 boys because the boy who got the first prize cannot receive second prize.

Similarly third prize can be given away to any of the remaining 2 boys

Hence total number of ways are 4 X 3 X 2 = 24

Note :- This is same as Arrangement of 4 boys taken 3 at a time in a way 4P3 = 4!/1! = 4! = 24 (More on this under the Head of Permutations later)

II) First prize to any one of the 4 boys. Similarly second to any one of the 4 boys, and third as well to any one of the 4 boys = 4 X 4 X 4 = 43 = 64

III) Since any one of the 4 boys may get all the prizes. So, the number of ways in which a boy gets all the 3 prizes = 4

So the number of ways in which a boy does not get all the prizes = 43 – 4 = 60

Permutations (Arrangement) nPr

Each of the arrangements which can be made by taking some or all of a number of things is called a permutation.

For example, if there are three objects namely x,y, and z,

then the permutations of these objects, taking two at a time, are xy, yx, yz, zy, xz, zx = Total 6 Permutations.

NOTE :- It should be noted that in permutations the order of arrangement is taken into account; When the order is changed, a different permutation is obtained.

Now Let’s go to some theoretical part