IBPS PO notification is out. From the official notification, we infer that the number of vacancies has gone down drastically. There will be a significant increase in the cut offs as the number of vacancies has halved. One of the topics that can help aspirants score is Permutation and combination for IBPS PO. Aspirants can try IBPS PO online preparation to aid them in the process.

Aspirants can check out previous year papers of IBPS PO to gauge the difficulty. As the competition is expected to increase by manifolds, aspirants must practice IBPS PO online mock tests to enhance their chances. Preparing for IBPS PO will give aspirants a shot at IBPS clerk exam too.

IBPS PO prelims exam consists of 3 sections – English language, quantitative ability and Reasoning ability. English language section contains 30 questions of 1 mark each while the other 2 sections contain 35 questions of 1 mark each.

## Permutation and combination for IBPS PO:

Among the 35 questions in quantitative aptitude section, 1 or 2 questions appear from Permutation and combination for IBPS PO. This topic is one of the most scoring since questions from this topic are not much calculation intensive.

In this article, let us discuss how to prepare Permutation and combination for IBPS PO.

Combination refers to the number of ways in which we can select a particular number of items from a given set of items. Permutations refer to the number of ways in which we can select a particular number of items and then arrange them.

The number of ways in which we can select Â â€˜xâ€™ items from â€˜nâ€™ items is same as the number of ways in which we can leave out â€˜n-xâ€™ items. Let us look at some problems to reinforce the concept.

### Question 1:

Find the number of ways in which 3 flowers can be selected from a basket of 5 flowers.

The number of ways in which we can select 3 flowers from 5 flowers Â = 5c3 = 10 ways. Aspirants must observe that this is same as rejecting 2 flowers ( 5c2).

### Question 2:

How many 3 lettered words can be using the five vowels?

One must be careful while solving such questions. The impulsive thing that most people do is to select 3 letters among the five vowels. However, the question does not mention that all the 3 letters are unique.

Hence, We can choose the first letter in 5 ways, the second letter in 5 ways and the third letter too in five ways. Total number of 3 lettered words = 5*5*5 = 125.

### Multiple counting:

### Question 3:

In how many ways can we arrange the letters in the word FREEDOM?

FREEDOM contains 2 Eâ€™s. If we simply assume the number of arrangements to be 7!, we will be counting each arrangement twice. Hence, we have to divide 7! by 2! to get the final answer.

Permutation is an extension of combination. After selecting the items, we have to arrange them. If there are â€˜nâ€™ items, we can arrange them in n! ways.

### Question 4:

How many 3 digit numbers can be formed such that all the digits are unique?

There are 10 unique single digit numbers – 0,1,2,3,4,5,6,7,8,and 9. From these, we can select the first digit in 9 ways ( 0 cannot be the first digit). We can select the second digit in 9 ways and the third digit in 8 ways. Hence, the total number of ways is 9*9*8 = 648.

Permutation and combination may also be required to solve some questions on probability.

### Question 5:

What is the probability that a randomly selected three lettered word that has unique letters has all its letters in alphabetical order?

We can arrange a 3 lettered word in 3! = 6 ways. Among these 6 possibilities, only 1 will have all its letters arranged in the right order. Hence, 1/6 is the correct answer.

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### Question 6:

What is the probability that a randomly selected 3 lettered word with unique letters has at least 1 vowel?

Instead of finding the probability of the word containing a vowel, let us find out the probability of the word containing no vowels and then subtract it from 1.

The total number of 3 lettered words with unique letters = 26*25*24.

Number of 3 lettered words (with unique letters) with no vowels = 21*20*19

The probability that the 3 lettered word does not contain a vowel is (21*20*19)/(26*25*24)= 7980/15600.

Hence, the probability that atleast one letter is a vowel is 1 – 7980/15600 = (15600 – 7980)/15600 = 7620/15600 = 381/780.

Questions from Permutation and combination for IBPS PO are quite simple and one can solve these questions easily. Aspirants must ensure that they score well in such topics to have a shot at getting their dream job.

Try reading our other useful blogs on IBPS PO profit loss questions and Input-output questions for IBPS PO.