If an event or a trial has ‘m’ outcomes and for every outcome of the previous event, a second event has ‘n’ outcomes, then the number of different ways by which both events can be conducted will be the product of m and n.
Consider the case of all outcomes possible if we flip 3 coins together.
Each toss results in one of 2 possible outcomes: heads or tails.
For each possible outcome of the first toss, there are two possible outcomes of the second toss. Thus, after the second toss, there are in all four possible outcomes 2 x 2 =4
For each of the four possible outcomes of the first and second tosses, there are two possible outcomes of the third toss. Thus, after the third toss, there are in all eight possible outcomes (2 x 2 x 2 =8). From the figure, all 8 combinations can be observed as HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT.
Permutations can also be termed as ordered choices or arrangements. Each of the arrangements that can be made by selecting ‘r’ things out of ‘n’ things can be termed as a permutation. In permutations, the order in which the items are arranged is significant.
Consider arranging r things selected from n things.
There are n possibilities for the first choice, (n – 1) possibilities for the second choice,(n – 2) possibilities for the third choice and so on. In other words, the available choices reduce by 1 after every selection. Therefore, ways to arrange r things selected out of n things are
n x (n-1) x (n-2) …….. r terms = n x (n-1) x (n-2)… (n-r+1)
As n! = n x (n-1) x (n-2) …….1, where the symbol ‘!’ represents the factorial function.
The above expression is the permutation function represented as
Consider a collection of 5 distinct items, out of which 3 have to be selected and arranged. They can be considered as a sequence of three choices, each of which results in the selection of one of the remaining things.
First choice – There are 5 possible outcomes.
Second choice – There will be 4 possible outcomes because 1 item has already been chosen. We don’t know what the possible outcomes are because they depend on the outcome of the first selection, but regardless of which item was picked first, 4 items remain as possibilities for the second choice.
Third choice– 3 possible outcomes, Therefore, to arrange 5 items in 3 places, the number of possible arrangements is-
Similar items in permutations
While arranging ‘n’ things from which ‘p’ things are of one kind and ‘q’ things are of the second kind, with the rest of the things being distinct, the number of different arrangements will be
and so on.
Repetition of items in permutations
The total number of ways in which ‘n’ things can be arranged in ‘r’ ways with repetition allowed are equal to nr ways.
When ‘n’ items are arranged in a circle, the number of arrangements of these ‘n’ items compared to arranging them in a straight line will be lesser. This happens because; in a straight line, we can have two different arrangements namely ABCD and DABC, whereas in a circle these two arrangements will be considered the same. To calculate the number of permutations of ‘n’ items arranged in a circle, one element can be fixed and all the various arrangement around it can be calculated. For example, to calculate the number of ways in which 5 women (Rose, Jean, Marie, Lily and Amy) can sit around a table, we can first arrange them in a straight line. There are 5! ways to arrange them.
The figure shows how some arrangements are different when arranged linearly and same when arranged in a circle.
Though the arrangements Rose-Jean-Lilly-Marie-Amy and Jean-Lilly-Marie-Amy-Rose are different, when arranged in a circular manner, they are same. If one of them, say, Lilly, is fixed, the number of ways in which the other women can sit around her is equal to 4! The number of ways in which ‘n’ distinct items can be arranged in a circle will be (n-1)!
Clockwise and Anti-clockwise arrangements
In certain cases, e.g., jewels in a necklace or garlands, there is no difference between clockwise and anticlockwise arrangements. In those cases, the total possible arrangements are half of the original ways of arrangements, i.e
Combinations can also be termed as selections. Hence, each of the selection of ‘r’ items made out of a set of ‘n’ items is called a combination. In combinations, the order in which items are arranged is not important.
2 students are to be selected from 10 students in a class for a scholarship. In this case, the order of the students in the pair is not important as they will anyway be given the same scholarship. Hence, this will be a combination or selection problem. In the given figure, selection of c followed by j is same as the selection of j followed by c for the scholarship. To find the number of combinations, the number of arrangements is calculated and is divided by 2! or 2.
Similarly, if we had to select 3 out of 10 students (named a, b, c…) for the scholarship, the total arrangements would be 10 x 9 x 8. Each way can be (say hij) can be arranged in 3 x 2 x 1 ways (hij, hji, ihj, ijh, jhi, jih). All the 6 arrangements select the same 3 students again. Therefore, to remove this duplication all arrangements are divided by 3!.
Generally, r things are selected from n things in
This is also represented as ncr.
It can be inferred that
Many problems can be simplified by breaking them down using this concept. If ‘m’ and ‘n’ are individual events, the total number of ways in which m and n occur together is mn. This has already been explained earlier. On the other hand, the total number of ways in which m or n may occur is (m + n).
To calculate the number of ways to fill two seats in a carriage from 20 men and 10 women when a man and a woman never travel together, we divide the problem in 2 cases.
Case1: When only men travel 20C2 = 190
Case 2: When only women travel 10C2 = 45
Case1 or Case 2 may occur, therefore, total possibilities are Case 1 + Case 2
i.e, 190 + 45 = 235
Source: Chalk Street
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