PNC is one of the most confusing topics for students. I have been told by students many times that they often get confused in understanding the selection vs arrangement logic while solving a permutation question. I assure you all, if you read this article till the very end, you will be able to handle most of the CUET PnC questions with ease, going forward.

Let’s understand the concepts by examples,

Q) Abhishek has his IIM K’s interview and he has come to a garment store to buy a dress for his interview. In the store he notices 20 shirts and 10 pants. In how many ways can Abhishek buy 1 shirt and 1 trouser?

Let’s see how to solve this basic question and understand the logic behind selection and arrangement,

As there are 20 shirts available. Let shirts be denoted by s1, s2, s3, s4 …… s20
As there are 10 pants available, Let pants be denoted by t1 , t2, t3, t4 ….. t10

As he has to buy 1 shirt and 1 pant and there is no preference available. So he can buy any shirt/pant combination

Any 1 shirt from the available 20 shirts and any 1 pant from the available 10 pants, The selection will be either of the following
s1t1, s1t2 , s1t3 ……… s1t10
s2t1, s2t2, s2t3, ……… s2t10
.
.
.
S20t1, s20t2, s20t3 …… s20t10

If you count these choices, you will see that they are 200 in number. This can be solved easily by, 20C1 (1 shirt to buy out of 20 choices) x 10C1 (1 pant to buy out of 10 choices) = 200

Note: In this question, only selection mattered. s1t5 is same as t1s1 (You buy 1st shirt with 5th pant, or 5th pant with 1st shirt, you will end up having the same combo)

General Rule 1 - If one operation can be performed in m ways and second operation can be performed in n ways than the number of ways of performing both the operation is (m x n)

Let’s see another question to validate our general rule,

Q) In how many ways can you post 10 letters in 4 letterboxes?

Before solving this question, visualize the problem statement. You have 10 letters with you and 4 letter boxes are there in front of you. The first letter can go in any letter box So, it has 4 choices. The second letter can again go in any letter box so it has 4 choices, similarly for 3rd, 4th … till 10th letters.

Now we know from the general rule that If one operation can be performed in m ways and second operation can be performed in n ways than the number of ways of performing both the operation is (m x n). In our case we have 10 operations. Each operation can be performed in 4 ways. So our answer would be 4^10

See the rule works like a magic. Can be pretty useful for CUET. But let's not be happy right now, we have a lot to learn

There are questions where arrangements also matter. Let’s see such kind of a question with an example,

Q) There are 5 people A, B, C, D, E. 3 out of these are to be seated in three chairs arranged in a line. In how many ways this can be done

Notice that here you have seats in which 3 people are to be seated. Now out of the 5 available people, lets take out A B and C randomly. These 3 people can occupy the three seats in,

| A B C | A C B | B A C | B C A | C A B | C B A |

You see, the way these same 3 people are seated would change the final outcome (sitting on the chair)

It can be seen from the above demonstration that any 3 people would have 6 arrangements.

The total number of ways by which 3 people can be selected out of 5 is 5C3 = 10 (ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE) And Each trio can be arranged in 6 ways (3!). So, the final answer would be 5C3*3! = 60. This can also be directly solved using 5P3 which is again 60.

General rule 2 – N number of people can be arranged in N! ways

Now, we know when to use permutation and combination and how they work. Let’s dive into some real questions that we can expect in CUET. But before starting, I want to share an advice which is, never to use P (P as in permutation) directly. Always break the question into small parts. Like, if arrangement is required, first select and then arrange (Like in the above question we did 5C3*3! instead of 5P3). This will make the solution more logical and there will be less chances of error.

Let's look at another question

Q) There are 5 boys and 2 girls. In how many ways can they be arranged in a line such that the 2 girls are always together.

Now we have an arrangement question but with a twist. There’s a condition involved. We need to arrange the people such that the 2 girls are always together.

In an ideal scenario, we have total 7 people. Using general rule 2, we can get 7! And this question would have been sorted. But, unfortunately that’s not the case as we have a condition to take care of.

Modification 1 to general rule 2:
As the 2 girls always need to be together, we can consider them a unit. Now, we would be having 5 boys and a unit. So total 6 things. If we arrange this set up, the 2 girls would always be together. The arrangement here can be done in 6! Ways. But wait, this is not our final answer. The unit has 2 girls. Recall our last arrangement question. In an arrangement set up order matters. The unit has G1 G2. Now when you are ordering 5 Boys and 1 Unit of (G1 G2) cases can be,

B1 B2 B3 G1 G2 B4 B5
B1 B2 B3 B4 B5 G1 G2
And so on………..,

Notice that G1 G2 can be arranged in every case within the unit which will make the arrangement different,
B1 B2 B3 G1 G2 B4 B5 and B1 B2 B3 G2 G1 B4 B5
B1 B2 B3 B4 B5 G1 G2 and B1 B2 B3 B4 B5 G2 G1
So, the final answer would be 6! X 2!

Let's see another question for a new set of modification,

Q) There are 5 boys and 2 girls. In how many ways can they be arranged in a line such that exactly 2 boys are always between the 2 girls

The question again got a little twisted. The condition has changed. Now, there need to be 2 boys between the 2 girls.

Modification 2 to general rule 2:
Few of the cases, from the total list of arrangements would look like this,
B1 B2 B3 G1 B4 B5 G2
B1 B2 G1 B3 B4 G2 B5

So basically, the boys between the 2 girls can change. Let’s do what we did earlier. Let’s create a unit of 2 girls and 2 boys by selecting the 2 boys first who needs to go between the girls. This can be done in 5C2 ways. Now we have 3 Boys and 1 unit having 2 boys between 2 girls, so 4 members.

B1 B2 B3 and G1 B4 B5 G2
Using general rule 2 , these 4 members can be arranged in 4! Ways ,

Now, we will have something like this
B1 B2 B3 (G1 B4 B5 G2)
B1 B2 (G1 B4 B5 G2) B3
And so on,

If you see every case, it is evident that in every case we can swap the position of the girls and can also swap the position of the boys between the girls.
Like the first case B1 B2 B3 G1 B4 B5 G2 can have the below arrangements within itself,

B1 B2 B3 G1 B4 B5 G2 | B1 B2 B3 G2 B4 B5 G1 | B1 B2 B3 G1 B5 B4 G2 | B1 B2 B3 G2 B5 B4 G1
So, the girls can be arranged in 2! Ways (G1 can take place of G2 and vice versa) and the boys between the girls can also be arranged in 2! Ways. Every arrangement in 4! selections can again be arranged in 2!*2! ways,

So the answer would be 5c2*4!*2!*2! = 960 ways

If you have read it till now, I am sure you have been intrigued. You must be wondering, is PnC this easy? Well at ExamDeewana it is. Let’s see the last rule before wrapping up,

Arrangement with duplicates:
Q) Arrange the letters AABCD to form all 4 letter words?

Let’s do this manually. We know we have 5 letters and we want to create all 4 letter words from those 5 letters. So definitely, we need to select 4 out of 5 first and then arrange them.So simple, right? Well not really. You see in this case if I do 5C4 I would get 5 pairs. These pairs would be AABC , AABD, AACD, ABCD, ABCD. Why 2 ABCDs? Well because we have 2 As. First A with BCD and second A with BCD

If you see the above combinations, we have a pair that is repeating. This means all the arrangements of the 2 ABCDs will be common. Thus. instead of 5C4 selections we would be using only 4 selections for arrangements. Woh! So much for the rules we learned above. Can we say that the number of selections are 2C2*3C2 + 1*3C3 = 4 (2As and any other 2 characters + 1A and all other 3 characters)

Now let’s see a case out of these 4 where repetition is taking place. Take for eg. AABC – If we arrange the letters of AABC we would get ABCA, ACBA, ABAC, ACAB, AABC, AACB, BCAA, CBAA, CABA, BACA, BAAC, CAAB, Thus only 12 arrangements

General rule 3 – If m objects have to arranged out of which n are same, then the possible number of arrangements is m!/n!

So, AABC can be arranged in 4!/2! = 12 ways AABD can be arranged in 12 ways AACD can be arranged in 12 ways And ABCD can be arranged in 24 ways (4! As nothing is common, General rule 2)

Total = 12 + 12 + 12 + 24 = 60

I am sure now you guys would be able to solve many such questions in CUET. For practice you can try the below mentioned questions which are roughly on the same concepts discussed above,

Practice 1)
Out of 11 members of a team two players has to be selected such that one is captain and another person is vice-captain. In how many ways it can be done.

Practice 2)
A village has 10 players. A team of 6 players is to be formed. 5 members are chosen first out of these 10 players and then the captain from the remaining players. Then the total number of ways of choosing such teams is

Practice 3)
How many numbers are there between 99 and 1000, having at least one of their digits 7?

Practice 4)
How many 5-digit telephone numbers can be constructed using the digits 0 to 9, if each number starts with 67 and no digit appears more than once?

Practice 5)
In a small village, there are 87 families, of which 52 families have at most 2 children. In a rural development programme, 20 families are to be chosen for assistance, of which at least 18 families must have at most 2 children. In how many ways can the choice be made?

Practice 6)
How many words can be made by arranging all the letters of the word MATHEMATICS

The author of this article is an alumni of IIM Kashipur and has scored a 99.45 percentile in CAT 2016’s quant section. For more details you can connect with him here > Linkedin

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