Statistics - Permutation and Combinations

This article discuss the use of permutations and combinations in finding the possibilities of arranging or numbe of combinations by using the permutation and combination techniques, particualrly in calculating theoritical probabilities of given events.
ntroduction
In Mathematics problems in permutations and combinations are very tricky. This is because ther are many problems which cannot be categorized in to a particualr method to be applied. The problems have to be solved by using unique insights and intuition and my experimenting with many ways of approch by using a combination of methods. However, by asking some systematic questions many permutations and combination problems can be solved. In this article I will discuss these questions and the categorise some of these problems which can be asked in High scholl Mathematics. This will certainly help students and people who are interested in Mathematics to explore these problems with a systematic way to approach such problems and enable to develop techniques for more complex problems.
The permutations and combinations can be categorised in to two major categories. They are Permutations and combinations without restrictions and with restrictions. It can be Line or circle arrangement in permutation questions. In these cases all items are chosen or some items are chosen. The items can be unique or identical. Restrictions can be categorized in to restricted items and unrestricted items or one or more restrictions. All permutations formulas are for the without replacement or repetition. The categories under without restriction and without replacement can be summariazed as follows:
Permutations and combinations without replacement has two major categories. They are as follows:
1. Without restrictions
2. With restrictions.
Without restriction and with restrictions can have permutations and combinations problems depending on whether order of arrangement is important or not. The categories under without restrictions and with restrictions are as follows:
1. Without restriction
a) Permutation
b) combination
2. With Restriction
a) Permutation
b) combination
Without restrictions, permutations can be firther categorized in to line arrangement or circle arrangement. Like wise with restriction category. This categorization is as follows:
1. Permutation - without restrictions
a) Line arrangement.
b) circle arrangement
2. Permutation with restriction
a) Line arrangement
b) circle arrangement
Line arrangement under permutation without restruction can be further catagorised as all items or some items.
This is as follows:
Permutation without restriction - Line arrangement
1. All items
2. Some items
Permutation without restriction, line arrangement of all items can be further catagorized in to each item unique or some items identical. This is as follows:
Permutation -Without replacement- Line arrangement - All items
1. Each item unique
2. some item identical.
Solutions for permutation and combination, where there is no restrictions in arrangement or chosing
1. Permutation without replacement and without restrictions line arrangement all items are unque = n factorial. Where n factorial = n*(n-1)8(n-2)......................................................1.
2. Permutation without replacement and without restrictions- Line arrangement, some items used = n p r. Where n p r = n factorial/ r factorial * (n-r) factorial.
3. Permutation without replacement and without restrictions, line arrangement some r items are identical and all items are chosen = n factorial/ r factorial.
4. Permutation without replacement, with restriction -Circle arrangement all items used = (n-1) factorial.
 Problems involving Permutations and combination without replacement and without restrictions
Question 1
How many ways 7 people be arranged in a line?
Answer 1
In this question it is an arrangement. There fore order important. That is it is a permutation problem involving all items without replacement and without restriction in a line arrangement. There fore the number of ways is equal to 7 factorial = 7*6*5*4*3*2*1= 5040.
Question 2
How many ways the letters of the word Wollongong be arranged?
Answer 2
This a line arrangement permutation where some items are identical. In this instant there are 10 letterers. where it has 3 letter o, 2 letters g and 2 letters l. Applying the formula for this category as per above solutions is as follows:
The ways of arrangement in line = 10 factorial/ 3 factorial*2factorial*2 factorial = 75600.
Question 3
How many different arrangement can be made using 3 letters from the word Monday?
Answer 3
This is a permutation question as it asks for order of arrangement. In this situation the letters are unique and without restrictions. As well, all letters are not arranges only some are arranged. There fore it is a permutation 3 letters out of 6 letters. That is 6 p 3 or 6*5*4 = 120.
Question 4
If people has to arranged ina round table how many different arrangements are there if there are 6 people?
Answer 4
This a circle arrnagement instead of line arrangement. There fore the the cicle permutation as per above solutions to the circle permutation without restrictions and without replacement have to be used. In this situation the number of arrangements are ( 6-1) factorial = 120.
Question 5
 How many committees can be formed out of 8 people?
Answer
In this question the order is not important. There fore it is not a permutation problem. It is a combination problem without replacement and without restrictions.  There fore applying the solutions formula as mentioned above is  8 c 4 = 8 factorial/ 4 factorial *( 8-4) factorial = 8*7*6*5/ 4*3*2*1 =  70
Solution Model for permutation and combination with restrictions
If the permutations and combinations have restrictions in the way the items have to placed or method of choosing then determine whether it is a line permutation or circle permutation or combination. After deciding from the problem the nature of the problem decide the patterns as per restrictions for a permutation problem of line arrangement and determine the number of arrangement of each restriction and then multiply it with the number of patterns. Say the number of patterns for a restricted permutation problem is (p) and the 1st restricted item of (u) is (u) factorial and the 2nd restriction is (r) items and there fore (r) factorial. Then the total number of restricted permutation = p*u factorial * r factorial.
If say the problem related to circle permutation, the leave one item out of the total number of items and use the line permutation with restrictions as discussed above. For example, say it has n items, ther fore one has to be taken away from the total items. That is (n-1) items are taken into consideration and then use the line arrangement model for the circle permutations.
If it is a combination problem, just like the permutation, the patterns or restrictions combinations have to calculated separately and then identify the different classes in accordance with the restriction or restriction rules.  Then multiply the combinations together with that of the number of combination necessary in consonant with the restriction to solve the problem.
In complex situations one has to calculate differently the permutations for different components of the problem and then add all the components together to get the final solutions. As well, in some situations combinations and permutations have to to multiplied to get the accurate results. These are exceptions to the general methodology adopted normally as discussed in this article.
Question 1
How many ways 4 boys and 3 girls can be arranged with 4 boys togeter in a line?
Answer 1
Restricted items: 4 boys
non- restricted items 3 girls
The possible patterns with restrictions are as follows:
BBBBggg
gbbbbgg
ggbbbbg
gggbbbb
That is there are 4 patterns. If no replacement then 4 boys can be arranged in 4 factorial ways, 3 boys can be arranged 3 factorial ways. If there are 4 patterns, then ther are 4*4 factorial * 3 factorial = 576.
Question 2
If  4 males and 3 females are there and they have to be arranged alternatively, how many arrangements are there?
Restricted items: 4 males and 3 females.
Restriction: they must be arranged alternatively that is m f m f m or f m f m. That is only one pattern
4 males can be arranged in 4 factorial ways. 3 females can be arranged 3 factorial ways, only one pattern. There fore the number of total arrangement = 1*4 factorial * 3 factorial = 144.
Question 3
In how many ways 5 adults can be arranged in a circle table, if say Bred and Biggs are not to sit near to each other?
Answer 3
This is a circle arrangement. Fix a position for Bred. Then consider the 4 adults as Biggs will not be at each end.
The possible patterns are
A B A A or A A B A
Bigg can be arranged in 1 factorial ways. The other 3 adults can be arranged in 3 factorial ways. there are two patterns. There fore the total arrangements with bred and Biggs are not sitting together is = 2* 1 factorial * 3 factorial = 12
Question 4
How many committees can be arranged if 3 males and 5 females have to be chosen from 6 males and 10 females?
Answer 4
This do not require order , there fore it is a combination problem with restriction that 3 males have to chosen from 6 males and  5 females from 10 females.
The number of combination if 3 males are chosen from 6 males = 6 C 3 = 6 factorial/ 3 factorial* 3 factorial = 20
The number of combinations of chosing 5 females from 10 females  = 10 c 5 = 10 factorial/ 5 factorial * 5 factorial = 252.
There fore, the number of committees are = 20*252 = 5040.
Question 5
In how many ways 5 adults can be arranged in a circle if Ben and Biggs must not sit near to each other?
Answer 5
This is a circle arrangement. Fix a position for Ben. The arrange the 4 adults in such a manner that Biggs do not sit at the end of a line arrangement. If this done, then there are two patterns. The are as follows:
A B A A
A A B A
Biggs can be arranged in 1 factorial way. Other 3 adults can be arranged in 3 factorial ways. There fore total arrangements in a circle with restriction = 2* 1 factorial * 3 factorial = 12.

Question 6
Calculate the probability if 4 males and 3 females are arranged in a line and they alternatively sit.

Answer 6
The number of arrangement possible if 4 males and 3 females are to be sitting alternatively = 1* 4 factorial * 3 factorial = 24 *6 =144

If no restrictions they can be arranged in 7 factorial ways = 7*6 * 5 * 4 *3 *2 *1 = 7*30* 24 = 240*24 = 5040
There fore the probability that the boys and girls alter nate arrangement = 144/ 5040 .