A saint has a magic pot. He puts one gold ball of radius 1 mm daily inside it for ten days. If the weight of the first ball is 1 gm and if the radius of a ball inside the pot doubles every day, how much gold has the saint made due to his magic pot?
 A)2^{30}  69/7
 B)2^{30} + 69/7
 C)2^{30}  71/7
 D)2^{30} + 71/7
i.e. the radius of the ball in the tenth day = 2^{9} mm
Sequence of weight per day = 1, 8, 64, 512,.... = 8^{0}, 8^{1}, 8^{2},..., 8^{9}
i.e. the weight of the ball in the tenth day = 8^{9} gm
Hint: If the radius = 1 mm → Volume = 4/3 π r^{3} = 4/3 π (1)^{3} = 1 (4/3 π) → Weight = 1 gm If the radius = 2 mm → Volume = 8(4/3 π) → Weight = 8 gm If the radius = 4 mm → Volume = 64(4/3 π) → Weight = 64 gm And so on. 
The ball put in the;  Weight of the ball on the 10th day 

1st day  89 gm 
2nd day  88 gm 
3rd day  87 gm 
.  . 
9th day  81 gm 
10th day  80 = 1 gm 
Hence the total weight of all the balls on 10^{th} day = 8^{0} + 8^{1} + 8^{2} + ..... + 8^{9}
= (8^{10}  1)/(81) = (2^{30}  1)/7 gm
The initial weight of 10 balls = 10 gm
Therefore, the gain in weight = [(2^{30}  1)/7]  10
= (2^{30}  1)/7  70/7
= (2^{30}  71)/7 gm
Ans: C
Fundamental rules of addition and multiplication
Rule 1: Addition principle
If there are 'n' independent events and the outcomes of each event is a, b, c, d..... respectively.
Then the total number of all the events = a + b + c + d +....
Example:
A box is containing 10 pens and 5 pencils. How many different ways you can select one pen OR one pencil from the box?
For selecting pen, there are 10 different ways.
For selecting pencil, there are 5 different ways.
For selecting one pen or one pencil, there are 10 + 5 = 15 ways possible.
In the above example the condition 'OR' has an important role, means 'OR' is dissolving the difference in identity of the product. i.e. either pen or pencil you have to select one product out of the total 15 products. Therefore there are 15 such possible ways.
Rule 2: Multiplication principle
If the number of events be 'm' and each of these 'm' events there are further 'n' events and which all are corresponding to the first set of 'm' events.
Then the number of ways to happen two events together is m * n.
Example:
A box is containing 10 pens and 5 pencils. How many different ways you can select one pen AND one pencil from the box?
For the selection of one pen, there are 10 different ways are possible.
Similarly for the selection of one pencil, there are 5 different possible ways.
Therefore the total number of possible ways to select one pen AND on pencil = 10 * 5 = 50 ways.
Example:
For reaching from City A to City B, there are 4 different roads and City C is connected with City B by 3 different roads. If a person needs to travel from City A to City C via City B, how many possible ways he can opt?
Number of ways from A to B = 4
Number of ways from B to C = 3
Total number of ways from A to C via B = 4 * 3 = 12
These 12 ways can be expressed in the form of ordered pairs, such as;
(a_{1}, b_{1}), (a_{1}, b_{2}), (a_{1}, b_{3})
(a_{2}, b_{1}), (a_{2}, b_{2}), (a_{2}, b_{3})
(a_{3}, b_{1}), (a_{3}, b_{2}), (a_{3}, b_{3})
(a_{4}, b_{1}), (a_{4}, b_{2}), (a_{4}, b_{3})
Result: An event E_{1} can be completed in 'm' different ways and another event E_{2} can be completed in 'n' different ways, then the number of ways of the completion of;

Basic concept of arrangement (Repetition of objects is allowable)
This is fundamental practice of arrangements. Here we are using the fundamental principle of multiplication and addition.
Example: How many five digit numbers are multiple of 5?
In this type of questions, we can apply the concept of place arrangements.
Divisibility of 5 depends upon the unit digit of the number, i.e. unit digit should be either 0 or 5.
Ten thousands place  Thousands place  Hundreds place  Tens place  Unit place 

1, 2, ...9  0, 1, 2, ...., 9  0, 1, 2, ...., 9  Same as thousands place  Same as ten thousands place 
9 ways  10 ways  10 ways  1 way  1 way 
Multiply all the possible ways to get the number of all the required numbers.
i.e. The number of required numbers = 9 * 10 * 10 * 10 * 2 = 18000
Conditional arrangements
This is the extension of the above concept. In this type we are considering some of the additional conditions in the arrangements, even though the repetition of the objects is allowable.
Example: How many 5 digits palindrome numbers are possible?
Palindrome numbers are numbers in which the extreme digits are same. While reading it from left to right or right to left, the value is same.
Ten thousands place  Thousands place  Hundreds place  Tens place  Unit place 

1, 2, ...9  0, 1, 2, ...., 9  0, 1, 2, ...., 9  Same as thousands place  Same as ten thousands place 
9 ways  10 ways  10 ways  1 way  1 way 
Hence the total number of required numbers = 9 * 10 * 10 * 1 * 1 = 900
Example: How many 4 letter codes can be formed such that codes should start and end with a vowel and middle two letters are consonants?
There are 5 vowels and 21 consonants in the alphabet.
So the possible counting can be considered in the following manner.
1st place  2nd Place  3rd place  4th Place 

5 ways  21 ways  21 ways  5 ways 
Therefore the total number of required codes = 5 * 21 * 21 * 5 = 11025