A collection of techniques and rules for counting the number of outcomes that can occur for a particular experiment can be used. They are called counting techniques.
Fundamental Principle of Counting
If the first operation can be performed in any one of the m ways and then a second operation can be performed in any one of the n ways, then both can be performed together in any one of the m*n ways.
This rule can be generalized. If first operation can be performed in any one of the n_{1} ways, second operation in any one of the n_{2} ways, ...... kth operation in any one of the nk ways, then together these can be performed in any one of the n_{1}× n_{2}× ...... × nk ways.
Permutation
A permutation is an arrangement of a given number of objects in a definite order.
(a) Permutations of n objects: The total number of permutations of n distinct objects is n!. Using symbols, we can write nPn= n!, (where n denotes the permutations of n objects, all taken together).
Let us assume there are n persons to be seated on n chairs. The first chair can be occupied by any one of the n persons and hence, there are n ways in which it can be occupied. Similarly, the second chair can be occupied in n  1 ways and so on. Using the fundamental principle of counting, the total number of ways in which n chairs can be occupied by n persons or the permutations of n objects taking all at a time is given by:
n_{Pn} = n(n – 1)(n – 2) ...... 3.2.1 = n!
(b) Permutations of n objects taking r at a time: In terms of the example, considered above, now we have n persons to be seated on r chairs, where r ≤ n.
Thus, n_{Pr} = n(n – 1)(n – 2) ...... [n – (r – 1)] = n(n – 1)(n – 2) ...... (n – r + 1).
On multiplication and division of the R.H.S. by (n  r)!, we get
(c) Permutations of n objects taking r at a time when any object may be repeated any number of times: Here, each of the r places can be filled in n ways. Therefore, total number of permutations is n_{r}.
(d) Permutations of n objects in a circular order: Suppose that there are three persons A, B and C, to be seated on the three chairs 1, 2 and 3, in a circular order. Then, the following three arrangements are identical:
Similarly, if n objects are seated in a circle, there will be n identical arrangements of the above type. Thus, in order to obtain distinct permutation of n objects in circular order we divide n_{Pn}by n, where n_{Pn}denotes number of permutations in a row.
Hence, the number of permutations in a circular ordern!/n = (n1)!
Permutations with restrictions: If out of n objects n_{1} are alike of one kind, n_{2} are alike of another kind, ...... nk are alike, the number of permutations are
n!/n_{1}!n_{2}!n_{3}!....n_{k}!
Since permutation of ni objects, which are alike, is only one (i = 1, 2, ...... k). Therefore, n! is to be divided by n_{1}!, n_{2}! .... n_{k}!, to get the required permutations.
Example : What is the total number of ways of simultaneous throwing of
(i) 3 coins,
(ii) 2 dice and
(iii) 2 coins and a die?
Solution:
(i) Each coin can be thrown in any one of the two ways, i.e, a head or a tail, therefore, the number of ways of simultaneous throwing of 3 coins = 2^{3} = 8.
(ii) Similarly, the total number of ways of simultaneous throwing of two dice is equal to 6^{2} = 36 and
(iii) The total number of ways of simultaneous throwing of 2 coins and a die is equal to 2^{2} * 6 = 24.
Example : A person can go from Delhi to PortBlair via Allahabad and Calcutta using following mode of transport :
In how many different ways the journey can be planned?
Solution: The journey from Delhi to PortBlair can be treated as three operations; From Delhi to Allahabad, from Allahabad to Calcutta and from Calcutta to PortBlair. Using the fundamental principle of counting, the journey can be planned in 4*4*2 = 32 ways.
Example : In how many ways the first, second and third prize can be given to 10 competitors?
Solution: There are 10 ways of giving first prize, nine ways of giving second prize and eight ways of giving third prize. Therefore, total no. ways is 10* 9* 8 = 720.
Alternative method: Here n = 10 and r = 3,
Example :
Solution:
Example : Three prizes are awarded each for getting more than 80% marks, 98% attendance and good behaviour in the college. In how many ways the prizes can be awarded if 15 students of the college are eligible for the three prizes?
Solution: Note that all the three prizes can be awarded to the same student. The prize for getting more than 80% marks can be awarded in 15 ways, prize for 90% attendance can be awarded in 15 ways and prize for good behaviour can also be awarded in 15 ways.
Thus, the total number of ways is n^{r} = 15^{3} = 3,375.
Example :
Solution:
Example :
Solution:
Example : There are 3 different books of economics, 4 different books of commerce and 5 different books of statistics. In how many ways these can be arranged on a shelf when
Solution:
Combination
When no attention is given to the order of arrangement of the selected objects, we get a combination. We know that the number of permutations of n objects taking r at a time is n_{Pr}. Since r objects can be arranged in r! ways, therefore, there are r! permutations corresponding to one combination. Thus, the number of combinations of n objects taking r at a time, denoted by n_{Cr}, can be obtained by dividing n_{Pr}by r!, i.e.,
Note:
Example :
Solution
Ordered Partitions
1. Ordered Partitions (distinguishable objects)
To illustrate this, let r = 3. Then n_{1} objects in the first compartment can be put in n_{Cn1} ways. Out of the remaining n – n_{1} objects, n_{2} objects can be put in the second compartment in n n_{1}_{Cn2} ways. Finally the remaining n – n_{1}– n_{2} = n_{3} objects can be put in the third compartment in one way. Thus, the required number of ways is
n_{cn1}*nn_{1}_{cn2}=n!/n_{1}!n_{2}!n_{3}!
2. Ordered Partitions (identical objects)
(a) The total number of ways of putting n identical objects into r compartments marked as 1, 2, ...... r, is n r 1_{Cr 1}, where each compartment may have none or any number of objects.
We can think of n objects being placed in a row and partitioned by the (r – 1) vertical lines into r compartments. This is equivalent to permutations of (n + r – 1) objects out of which n are of one type and (r – 1) of another type.
The required number of permutations are ( n+r1/n!(r1)), which is equal to (n+r1)_{Cn}or(n+r1)_{Cr1}
(b) The total number of ways of putting n identical objects into r compartments is (nr)+(r1)_{C(r1)}or(n1)_{C(r1)} , where each compartment must have at least one object.
In order that each compartment must have at least one object, we first put one object in each of the r compartments. Then the remaining (n – r) objects can be placed as in (a) above.
(c) The formula, given in (b) above, can be generalised. If each compartment is supposed to have at least k objects, the total number of ways is(nkr )+ (r1 )_{C(r1)} where k = 0, 1, 2, .... etc. such that K=n/r
Example : 4 couples occupy eight seats in a row at random. What is the probability that all the ladies are sitting next to each other?
Solution: Eight persons can be seated in a row in 8! ways.
We can treat 4 ladies as one person. Then, five persons can be seated in a row in 5! ways. Further, 4 ladies can be seated among themselves in 4! ways.
∴ The required probability = 5!4!/ 8! = 1/14
Example: 12 persons are seated at random (i) in a row, (ii) in a ring. Find the probabilities that three particular persons are sitting together.
Solution:
Example : 5 red and 2 black balls, each of different sizes, are randomly laid down in a row. Find the probability that
Solution: The seven balls can be placed in a row in 7! ways.
Example: Each of the two players, A and B, get 26 cards at random. Find the probability that each player has an equal number of red and black cards.
Solution: Each player can get 26 cards at random in 52_{C26} ways.
In order that a player gets an equal number of red and black cards, he should have 13 cards of each colour, note that there are 26 red cards and 26 black cards in a pack of playing cards. This can be done in 26_{C13} , 26_{C13} ways.
Hence, the required probability = 26_{C13} x 26_{C13} / 52_{C26}
Example : 8 distinguishable marbles are distributed at random into 3 boxes marked as 1, 2 and 3. Find the probability that they contain 3, 4 and 1 marbles respectively.
Solution: Since the first, second.... 8th marble, each, can go to any of the three boxes in 3 ways, the total number of ways of putting 8 distinguishable marbles into three boxes is 38.
The number of ways of putting the marbles, so that the first box contains 3 marbles, second contains 4 and the third contains 1, are 8!/ 3!4!1!
∴ The required probability = 8!/ 3!4!1! x 1/38 = 280/ 6561.
Example: 12 'one rupee' coins are distributed at random among 5 beggars A, B, C, D and E. Find the probability that:
(i) They get 4, 2, 0, 5 and 1 coins respectively.
(ii) Each beggar gets at least two coins.
(iii) None of them goes empty handed.


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Quantitative Techniques For Management Tutorial
Quantitative Techniques – Introduction
Measures Of Central Tendency
Mathematical Model
Linear Programming: Graphical Method
Linear Programming: Simplex Method
Transportation Model
Assignment Model
Network Model
Waiting Model (queuing Theory)
Probability
Theoretical Probability Distributions
Probability Distribution Of A Random Variable
Inventory Model
Game Theory
Simulation
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