FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 2.434! 4 3 2! 4 3 12
2! 2!
= × × = × =
(ii) The relative order of vowels and consonants unaltered means that vowel will take place of vowel
and consonant will take place of consonant. Now the 3 vowels can be arranged among themselves
in 3! ways, while 4 consonants with 2P can be arranged in
4! 4 3 2! 4 3 12
2! 2!
= × × = × =
ways.
So total number of ways of rearrangement in which the given arrangement is included
= 3! × 12 = 6 × 12 = 72
Hence, Required number of arrangement = 72 – 1 = 71.
Example 70 : How many numbers between 5000 and 6000 can be formed with the digits 3, 4, 5, 6, 7, 8?
Solution :
The number to be formed will be of 4 figures, further digit 5 is to be placed in 1st place (from left). Now the
remaining 3 places can be filled up by the remaining 5 digits in^5 P 3 ways.
Hence, required no.^53
P 1^5! 60
= × =2!=
Example 71 : In how many ways can be letters of the word SUNDAY be arranged? How many of them do
not begin with S? How many of them do not begin with S, but end with Y?
There are 6 letters in the word SUNDAY, which can be arranged in 6! = 720 ways.
Now placing S in first position fixed, the other 5 letters can be arrange in (5)! = 120 ways.
The arrangements of letters that do not begin with S = (6)! – (5)! = 720 – 120 = 600 ways.
Lastly, placing Y in the last position, we can arrange in (5)! = 120 ways and keeping Y in the last position and
S in the first position, we can arrange in (4)! = 24 ways.
Hence, the required no. of arrangements = (5)! – 4! = 120 – 24 = 96 ways.
(Problems regarding ring or circle)
Example 72 : In how many ways 8 boys can form a ring?
Solution :
Keeping one boy fixed in any position, remaining 7 boys can be arranged in 7! ways.
Hence, the required on. of ways = 7! = 7. 6. 5. 4. 3. 2. 1 = 5040.
Example 73: In how many ways 8 different beads can be placed in necklace?
Solution :
8 beads can be arranged in 7! ways. In this 7! ways, arrangements counting from clockwise and anticlockwise
are taken different. But necklace obtained by clockwise permutation will be same as that obtained from
anticlockwise. So total arrangement will be half of 7 !.
Hence, required no. of ways = ½ × 7! = ½ × 5040 = 2520.
Example 74: In how many ways 5 boys and 5 girls can take their seats in a round table, so that no two girls
will sit side by side.