{"id":7147,"date":"2022-04-15T17:47:46","date_gmt":"2022-04-15T12:17:46","guid":{"rendered":"https:\/\/www.goseeko.com\/blog\/?p=7147"},"modified":"2025-12-27T11:18:01","modified_gmt":"2025-12-27T05:48:01","slug":"what-is-permutation-and-combination","status":"publish","type":"post","link":"https:\/\/www.goseeko.com\/blog\/what-is-permutation-and-combination\/","title":{"rendered":"What is permutation and combination?"},"content":{"rendered":"\n

Overview(permutation and combination)<\/h2>\n\n\n\n

Before studying about permutation and combination<\/a>, we will have to know about factorial. <\/p>\n\n\n\n

Here we denote The product of the positive integers from 1 to n by n! And we read it as \u201cfactorial \u201c<\/p>\n\n\n\n

\u201cThe continued product of first n natural numbers, i.e., 1, 2, 3, \u2026. (n \u2013 1) n, we generally express this by the symbol n! <\/p>\n\n\n\n

we read it as \u2018factorial n\u201d.\u201d therefore, it is expressed as below<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Hence, using the symbol of permutation, we get,<\/p>\n\n\n\n

nPr = n (n \u2013 1) (n \u2013 2) \u2026 (n \u2013 r + 1)<\/p>\n\n\n\n

Example: Get the value of the following factorials<\/p>\n\n\n\n

1.     5!,<\/p>\n\n\n\n

2.      6! \u2212 <\/em>5! <\/p>\n\n\n\n

Sol.<\/p>\n\n\n\n

(i) 5! = 5 \u00d7 <\/em>4 \u00d7 <\/em>3 \u00d7 <\/em>2 \u00d7 <\/em>1 = 120.<\/p>\n\n\n\n

(ii) 6! \u2212 <\/em>5! = 6 \u00d7 <\/em>5! \u2212 <\/em>5! = (6 \u2212 <\/em>1) \u00d7 <\/em>5! = 5 \u00d7 <\/em>120 = 600.<\/p>\n\n\n\n

 Fundamental principal of counting<\/h2>\n\n\n\n
\"\"<\/figure>\n\n\n\n

The Sum Rule<\/strong> of permutation and combination<\/strong><\/h2>\n\n\n\n

Let us consider two tasks which need to be completed. <\/p>\n\n\n\n

If the first task can be completed in M <\/em>different ways and the second in N <\/em>different ways, and if these cannot be performed simultaneously, then there are M <\/em>+ N <\/em>ways of doing either task. above all, This is the sum rule of counting.<\/p>\n\n\n\n

The Product Rule of permutation and combination<\/strong><\/h2>\n\n\n\n

Let’s suppose that a task comprises of two procedures. If the first procedure can be completed in M <\/em>different ways and the second procedure can be done in N <\/em>different ways after the first procedure is done, then the total number of ways of completing <\/em>the task is M \u00d7 N<\/em><\/p>\n\n\n\n

Example<\/strong><\/h2>\n\n\n\n

Suppose one woman or one man has to be selected for a game from a society comprising 17 men and 29 women. <\/strong><\/p>\n\n\n\n

In how many different ways can this selection be made?<\/strong><\/p>\n\n\n\n

Sol.<\/p>\n\n\n\n

The first task of selecting a woman can be done in total 29 ways. <\/p>\n\n\n\n

The second task of selecting a man can be done in 17 ways. It follows from the sum rule, that there are 17+29 = 46 ways of making this selection.<\/p>\n\n\n\n

Permutations<\/a><\/h2>\n\n\n\n

\u201cThe different arrangements which can be made out of a given set of things, by taking some or all of them at a time are called permutations.\u201d<\/p>\n\n\n\n

Or<\/p>\n\n\n\n

A permutation is an arrangement of objects in a definite order. Since we have already studied combinations, we can also interpret Permutations as \u2018ordered combinations.<\/p>\n\n\n\n

A permutation of n objects taken r at a time is an arrangement of r of the objects<\/p>\n\n\n\n

(r\u2264n).<\/p>\n\n\n\n

The symbols of permutations of n things taken r at a time are-  <\/p>\n\n\n\n

Note<\/h2>\n\n\n\n

1.     The number of permutations of n different things taken r at a time in which p particular things never occur is-<\/p>\n\n\n\n

2.     The number of permutations of n different things taken r at a time in which p particular things are always present is-<\/p>\n\n\n\n

Note- A permutation of n <\/em>objects taken r <\/em>at a time is also called r<\/em>-permutation<\/p>\n\n\n\n

Permutation<\/h2>\n\n\n\n

The arrangement of a set of n objects in a given order is called a permutation.<\/p>\n\n\n\n

Any arrangement of any of these objects in a given order is said to be r-permutation.<\/p>\n\n\n\n

Therefore, The number of permutations of n objects taken r will be denoted as-<\/p>\n\n\n\n

P(n, r)<\/p>\n\n\n\n

Formula-                        <\/p>\n\n\n\n

Permutation with repetitions<\/h2>\n\n\n\n

The number of permutations of n objects of which are q_1 are alike, q_2are alike, q_r<\/p>\n\n\n\n

are alike is-<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Sampling with or without replacement<\/h2>\n\n\n\n

Suppose we chose the samples with repetition, from example if we draw a ball from a urn then we put back that ball in the urn and again we pick a ball and we continue the process.<\/p>\n\n\n\n

So this is the case of sampling with repetition then the product rule tells us that the number of such samples is-<\/p>\n\n\n\n

n.n.n.n.n\u2026\u2026\u2026.n.n =n^r<\/p>\n\n\n\n

And if we pick a ball from the urn and we do not put it back to the urn, then this is the case of sampling without replacement.<\/p>\n\n\n\n

Therefore, in this case the number of samples are-<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Example<\/h2>\n\n\n\n

If i choose three cards one after the other from a 52-card deck. Find the number m of ways<\/p>\n\n\n\n

We can do this in: (a) with replacement; (b) without replacement.<\/p>\n\n\n\n

Sol.<\/p>\n\n\n\n

(a) Each card can be chosen in 52 ways, so that, m = 52(52)(52) = 140 608.<\/p>\n\n\n\n

(b) Here there is no replacement. he first card can be chosen in 52 ways, the second in 51 ways, and the third in 50 ways. So that<\/p>\n\n\n\n

P(52, 3) = 52(51)(50) = 132 600<\/p>\n\n\n\n

Example- There are 4 black, 3 green and 5 red balls. In how many ways we can arrange in a row?<\/strong><\/p>\n\n\n\n

Solution: Total number of balls = 4 black + 3 green + 5 red = 12<\/p>\n\n\n\n

The black balls are alike,<\/p>\n\n\n\n

The green balls are, and the red balls are alike,<\/p>\n\n\n\n

Hence, the number of ways in which we can arrange the balls in a row =<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Example<\/h2>\n\n\n\n

A box contains 10 light bulbs. Find the number n of ordered samples of:<\/p>\n\n\n\n

(a) Size 3 with replacement,<\/p>\n\n\n\n

(b) Size 3 without replacement.<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Example<\/h2>\n\n\n\n

In a sports broadcasting company, the manager must pick the top three goals of the month, from a list of ten goals. In how many ways we can select the top three goals?<\/p>\n\n\n\n

Solution: Since the manager must decide the top three goals of the month, the order of the goals is very important! It decides the first-place winner, the runner-up, and the second runner-up. Thus, we can see that the problem is of permutation formula.<\/p>\n\n\n\n

Picking up three goals from a list of ten:<\/p>\n\n\n\n

Possible Permutations = <\/p>\n\n\n\n

Therefore, there are 720 ways of picking the top three goals!<\/p>\n\n\n\n

Circular permutations<\/strong><\/h2>\n\n\n\n

A circular permutation of n <\/em>objects is an arrangement of the objects around a circle.<\/p>\n\n\n\n

If the n <\/em>objects to be arrang round a circle we take an objects and fix it in one position.<\/p>\n\n\n\n

Now the remaining (n <\/em>\u2013 1) objects can be arranged to fill the (n <\/em>\u2013 1) positions the circle in (n <\/em>\u2013 1)! ways.<\/p>\n\n\n\n

Hence the number of circular permutations of n <\/em>different objects = (n <\/em>\u2013 1)!<\/p>\n\n\n\n

Example: In how many ways 8 girls can form a ring?<\/strong><\/p>\n\n\n\n

Sol.<\/p>\n\n\n\n

If we keep one girl fixed in any position, remaining 7 girls, we can arrange them in 7! ways.<\/p>\n\n\n\n

Hence, the required on. of ways <\/p>\n\n\n\n

= 7 ! = 7. 6. 5. 4. 3. 2. 1 = 5040.<\/p>\n\n\n\n

Combination<\/a><\/h2>\n\n\n\n

A combination of n objects taken at a time is an unordered selection of r of the n<\/p>\n\n\n\n

Objects (r \u2264n).<\/p>\n\n\n\n

\u201cThe different groups or collection or selections that can be made of a given set of things by taking some or all of them at a time without any regard to the order of their arrangements are called their combinations.\u201d<\/p>\n\n\n\n

Here note that a combination of n objects taken r at a time is also called r-combination of n objects.<\/p>\n\n\n\n

Therefore, The number of combinations are as-<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Total number of combinations<\/h2>\n\n\n\n

Total number of combination of n different things taken 1, 2, 3 \u2026. n at a time<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Here note that-<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Grouping<\/h2>\n\n\n\n

When we are require to form two groups out of (m + n) things, so that one group consists of m things and the other of n things. Now formation of one group represents the formation of the other group automatically. <\/p>\n\n\n\n

Hence, the number of ways m things we can select from (m + n) things.<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Example: If we need to choose for a competition from a class of 3 students, 4 students, in how many ways we can select them?<\/strong><\/p>\n\n\n\n

Sol.<\/p>\n\n\n\n

The number of combinations will be-<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

therefore, these are the all ways that we can select them.<\/p>\n\n\n\n

Interested in learning about similar topics? Here are a few hand-picked blogs for you!<\/p>\n\n\n\n