If the arrangements of objects are taken in circular order instead of a line then it is known as a circular permutation. For example, the arrangements of people in a round table. In such case
Example:In how many ways can 8 students be seated in a circle and in a line?
Solution:
The 8 students can be seated in a circle in (8-1)! = 7! =5070 ways
The 8 students can be seated in a line in 8! = 40320 ways
Restricted Permutation:
A Restricted permutation is a special type of permutation in which certain types of objects or data are always included or excluded and if they can come together or always stay apart.
(a)Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is to be always included in each arrangement
= r ^{n-1} P_{r-1}
(b)Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is fixed: = ^{n-1} P_{r-1}
(c) TheNumber of permutations of ‘n’ things, taken ‘r’ at a time when a particular thing is never taken: = ^{n-1} P_{r.}
(d) TheNumber of permutations of ‘n’ things, taken ‘r’ at a time, when ‘m’ specified things always come together = m! x ( n-m+1) !
(e) TheNumber of permutations of ‘n’ things, taken all at a time, when ‘m’ specified things always come together = n ! - [ m! x (n-m+1)! ]
Example: How many words can be formed with the letters of the word ‘NOTES’ when:
(i)‘N’ and ‘S’ occupying end places.
(ii)‘E’ being always in the middle
(iii)Vowels occupying odd-places
(iv)Vowels being never together.
Ans.
(i) When ‘N’ and ‘S’ occupying end-places
O.T.S (NS)
Here (NS) are fixed, hence O, T, S can be arranged in 3! ways
But (NS or SN) can be arranged themselves is 2! ways.
Total number of words = 3! x 2! = 12 ways.
(ii)When ‘E’ is fixed in the middle N.O.(E),T.S.
Hence four-letter N.O.T.S can be arranged in 4! i.e 24 ways.
(iii)Two vowels (O,E,) can be arranged in the odd places (1^{st}, 3^{rd})OR (3^{rd}^{,}5^{th}) or (1st,5th)= 2! x 2! x 2! ways = 8 ways
And three consonants (N,T,S) can be arranged in the even place (2^{nd}, 4^{th}) = 2 ! ways
The total number of ways=8 x2!=16 ways.
(iv)Total number of words = 5! = 120!
If all the vowels come together, then we have: (O.E.),N,T,S
These can be arranged in 4! ways.
But (O,E.) can be arranged themselves in 2! ways.
Number of ways, when vowels come together = 4! x 2!= 28 ways
The number of ways, when vowels being never-together= 120-28 = 92 ways.
Misclleceneous Problem of Circular and Restricted permutation:
Example:Find the number of ways in which 5 people A,B,C,D,E can be seated at a round table, such that
(i) A and B must always sit together.
(ii) C and D must not sit together.
Soln. (i) If we wish to set A and B together in all arrangements, we can consider these two as one unit, along with 3 others. So effectively we’ve to arrange 4 people in a circle, the number of ways being (4 – 1)! or 6.
But in each of these arrangements, A and B can themselves interchange places in 2 ways.
Therefore, the total number of ways will be 6 x 2 = 12.
(ii) The number of ways, in this case would be obtained by removing all those cases (from the total possible) in which C & D are together. The total number of ways will be (5 – 1)! or 24. Similar to (i) above, the number of cases in which C & D are seated together will be 12. Therefore the required number of ways will be 24 – 12 = 12.
Example:
In how many ways can 3 men and 3 ladies be seated at around table such that no two men are seated together?
Soln. Since we don’t want the men to be seated together, the only way to do this is to make the men and women sit alternately. We’ll first set the 3 women, on alternate seats, which can be done in (3 – 1)! or 2 ways.(We’re ignoring the other 3 seats for now. If each of the women is shifted by a seat in any direction, the seating arrangement remains exactly the same. That is why we have only 2 arrangements.
Now that we’ve done this, the 3 men can be seated in the remaining seats in 3! or 6 ways. Note that we haven’t used the formula for circular arrangements now. This is so because, after the women are seated, shifting the each of the men by 2 seats, will give a different arrangement. After fixing the position of the women (same as ‘numbering’ the seats), the arrangement of the remaining seats is equivalent to a linear arrangement.
Therefore the total number of ways, in this case will be 2! X 3! = 12.
Taken reference from
( Basic mathematics Grade XII and A foundation of Mathematics Volume II and Wikipedia.com )
A Restricted permutation is a special type of permutation in which certain types of objects or data are always included or excluded and if they can come together or always stay apart.
Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is to be always included in each arrangement
= r ^{n-1} P_{r-1}
Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is fixed: = ^{n-1} P_{r-1}
TheNumber of permutations of ‘n’ things, taken ‘r’ at a timewhen a particular thing is never taken: = ^{n-1} P_{r.}
TheNumber of permutations of ‘n’ things, taken ‘r’ at a time, when ‘m’ specified things always come together = m! x ( n-m+1) !
TheNumber of permutations of ‘n’ things, taken all at a time, when ‘m’ specified things always come together = n ! - [ m! x (n-m+1)! ]
ASK ANY QUESTION ON Circular, Restricted permutation
No discussion on this note yet. Be first to comment on this note