Page 87 - DMTH404_STATISTICS

Unit 6: Probability

To illustrate this, let r = 3. Then n objects in the first compartment can be put in Notes
n 1
C ways. Out of the remaining n - n objects, n objects can be put in the second
n
2
1
1
compartment in n- n 1 C ways. Finally the remaining n - n - n = n objects can be put
n
3
2
1
2
in the third compartment in one way. Thus, the required number of ways is
! n
n n- 1 n .
C  C =
1 n 2 n n !n !n !
1 2 3
2. Ordered Partitions (identical objects)
(a) The total number of ways of putting n identical objects into r compartments marked
n+r- 1
as 1, 2, ...... r, is C , where each compartment may have none or any number
r- 1
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
(n r+ - ) 1 ! (n r+ - ) 1 (n r+ - ) 1
permutations are , which is equal to C or C (r- ) 1 .
n
( ! n r - ) 1 !
(b) The total number of ways of putting n identical objects into r compartments is
(n r- ) (r+ - ) 1 (n- ) 1
C or C , where each compartment must have at least one object.
(r- ) 1 (r- ) 1
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 (n kr- ) (r+ - ) 1 C , where k = 0, 1, 2,
(r- ) 1
n
.... etc. such that k < .
r

Example 17: 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.

5!4! 1
 The required probability = = .
8! 14

Example 18: 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.

10!3! 1
(i) The required probability = = .
12! 22

9!3! 3
(ii) The required probability = = .
11! 55

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