Page 84 - DMTH404_STATISTICS

Statistics

Notes
Example 12: 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.
3
r
Thus, the total number of ways is n = 15 = 3,375.
Example 13:
(a) In how many ways can the letters of the word EDUCATION be arranged?
(b) In how many ways can the letters of the word STATISTICS be arranged?
(c) In how many ways can 20 students be allotted to 4 tutorial groups of 4, 5, 5 and 6 students
respectively?
(d) In how many ways 10 members of a committee can be seated at a round table if (i) they can
sit anywhere (ii) president and secretary must not sit next to each other?
Solution.
(a) The given word EDUCATION has 9 letters. Therefore, number of permutations of 9 letters
is 9! = 3,62,880.
(b) The word STATISTICS has 10 letters in which there are 3S' , 3T' , 2I' , 1A and 1C. Thus, the
s
s
s
10!
required number of permutations = 50,400.
3!3!2!1!1!
20!
(c) Required number of permutations = 9,77,72,87,522
4!5!5!6!
(d) (i) Number of permutations when they can sit anywhere = (10-1)!= 9! = 3,62,880.

(ii) We first find the number of permutations when president and secretary must sit
together. For this we consider president and secretary as one person. Thus, the
number of permutations of 9 persons at round table = 8! = 40,320.
 The number of permutations when president and secretary must not sit together =
3,62,880 - 40,320 = 3,22,560.

Example 14:
(a) In how many ways 4 men and 3 women can be seated in a row such that women occupy the
even places?
(b) In how many ways 4 men and 4 women can be seated such that men and women occupy
alternative places?
Solution.
(a) 4 men can be seated in 4! ways and 3 women can be seated in 3! ways. Since each arrangement
of men is associated with each arrangement of women, therefore, the required number of
permutations = 4! 3! = 144.

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