Page 261 - DCOM203_DMGT204_QUANTITATIVE_TECHNIQUES_I

Quantitative Techniques – I

Notes (c) The formula, given in (b) above, can be generalised. If each compartment is supposed
n kr r 1
to have at least k objects, the total number of ways is C r 1 , where k = 0, 1, 2,
n
.... etc. such that k .
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.

5!4! 1
The required probability
8! 14

Example: 12 persons are seated at random (1) in a row, (2) in a ring. Find the probabilities
that three particular persons are sitting together.
Solution:

10!3! 1
1. The required probability
12! 22
9!3! 3
2. The required probability
11! 55

Example: 5 red and 2 black balls, each of different sizes, are randomly laid down in a
row. Find the probability that
1. the two end balls are black,
2. there are three red balls between two black balls and
3. the two black balls are placed side by side.

Solution:
The seven balls can be placed in a row in 7! ways.
1. The black can be placed at the ends in 2! ways and, in-between them, 5 red balls can be
placed in 5! ways.
2!5! 1
The required probability
7! 21
2. We can treat BRRRB as one ball. Therefore, this ball along with the remaining two balls
can be arranged in 3! ways. The sequence BRRRB can be arranged in 2! 3! ways and the three
red balls of the sequence can be obtained from 5 balls in ways.

3!2!3! 5 1
The required probability C 3
7! 7

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