If you’re preparing for competitive exams, quantitative aptitude and permutations and combinations are two closely related topics that are essential to revise. Permutations and combinations are two important concepts in mathematics that are frequently tested in quantitative aptitude exams. A permutation is an arrangement of objects from a given set into a specific order, while a combination is a selection of objects from a given set in any order. Understanding the difference between the two can be tricky, but by learning some quick tips for solving problems related to permutations and combinations, it can be made easier.

We have already discussed this topic in detail in this article. Now, let’s focus on practice questions related to permutations and combinations in the context of Quantitative Aptitude. There are different types of questions that you can use to test your understanding, so start revising!

**Practice Quiz:**

## Permutation:

A Permutation is a collection or a combination of objects from a set where the order or the arrangement of the chosen objects does matter. In other words, a Permutation is an arrangement of objects in a definite order, For example, if we have two elements A and B, then there are two possible arrangements, ( A B ) and ( B A ).

**Formulas for Permutations:**

**Formulas for Permutations:**

^{n}P_{n}= n(n – 1) (n -2)Â^{n}P_{0 }= 1^{n}P_{1}= n_{Â }^{n}P_{n-1}= n!^{n}P_{r }= n.^{n-1}P_{r-1 }= n(n-1)^{n-2}P_{r-2}

The number of permutations when ‘

‘ elements are arranged out of a total of ‘n’ elements isr^{n}P_{r}Â= n! / (n – r)!For example, let n = 4 (A, B, C and D) and r = 2 (All permutations of size 2). The answer is 4!/(4-2)! = 12. The twelve permutations are AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB, and DC.

**Combination:**

**Combination:**

A** Combination** is the different

**of a given number of elements taken one by one, or some, or all at a time. For example, if we have two elements**

**selections****and**

**A****, then there is only one way to select two items, we select**

**B****of them.**

**both****Formulas for Combinations:**

**Formulas for Combinations:**

^{n}C_{râ€‹}is a natural number^{n}C_{0}â€‹=(^{n}C_{n}â€‹)=1^{n}C_{1}â€‹=n^{n}C_{r}â€‹=(^{n}C_{nâˆ’r}â€‹)^{n}C_{xâ€‹}=^{n}C_{y}â€‹ Â â‡’x=y or x+y=n- n.
^{nâˆ’1}C_{râˆ’1â€‹}=(nâˆ’r+1)Ã—^{n}C_{râˆ’1â€‹}

Â

The number of combinations when ‘r’ elements are selected out of a total of ‘n’ elements is

^{n}C_{r}= n! / ((r !) x (n – r)!).For example, let n = 4 (A, B, C and D) and r = 2 (All combinations of size 2). The answer is 4!/((4-2)!*2!) = 6. The six combinations are AB, AC, AD, BC, BD, and CD.

** Note:** In the same example, we have different cases for permutation and combination. For permutation, AB and BA are two different things but for selection, AB and BA are the same.

## Sample Questions on Permutations and Combinations

**Q1: **How many words can be formed by using 3 letters from the word **“DELHI”**?Â

**Q1:**

**“DELHI”**

**Solution:Â **

We need toÂ use combinations, not permutations. Combinations refer to the selection of items without regard to their order, whereas permutations take order into account. The formula for combinations is: nCr = n! / (r! * (n – r)!), where n is the total number of items, r is the number of items we want to choose. Using the formula for combinations: nCr = 5! / (3! * (5 – 3)!) nCr = 5! / (3! * 2!) nCr = (5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * (2 * 1)) nCr = (120) / (6 * 2) nCr = 120 / 12 nCr = 10 So, there are 10 different words that can be formed by using 3 letters from the word “DELHI.”

**Q2: **How many words can be formed by using the letters from the word **“DRIVER”** such that all the vowels are always together?Â

**Q2:**

**“DRIVER”**

**Solution:Â **

In these types of questions, we assume all the vowels to be a single character, i.e., “IE” is a single character. So, now we have 5 characters in the word, namely,

, andD, R, V, RBut,IE.occursR. => Number of possible arrangements =2 times=5! / 2!Now, the two vowels can be arranged in60= 2 ways. => Total number of possible words such that the vowels are always together=2!=60 x 2Â120

**Q3: **In how many ways, can we select a team of 4 students from a given choice of 15?

**Q3:**

**Solution:Â **

Number of possible ways of selection =

^{15}C=_{4}15 ! / ((4 !) x (11 !))

Number of possible ways of selection ==(15 x 14 x 13 x 12) / (4 x 3 x 2 x 1)1365Â

**Question 4: **In how many ways can a group of 5 members be formed by selecting **3** boys out of **6 boys** and **2** girls out of** 5 girls**?

**Question 4:**

**3**

**6 boys**

**2**

**5 girls**

**Solution:Â **

Number of ways 3 boys can be selected out of 6 =

^{6}C_{3}= 6 ! / [(3 !) x (3 !)] = (6 x 5 x 4) / (3 x 2 x 1) = 20 Number of ways 2 girls can be selected out of 5 =^{5}C_{2}= 5 ! / [(2 !) x (3 !)] = (5 x 4) / (2 x 1) = 10 Therefore, total number of ways of forming the group = 20 x 10 = 200Â

**Question 5: **How many words can be formed by using the letters from the word **“DRIVER”** such that all the vowels are never together?

**Question 5:**

**“DRIVER”**

**Solution:Â **

we assume all the vowels to be a single character, i.e.,

is a single character. So, now we have 5 characters in the word, namely,“IE”andD, R, V, R,. But, R occurs 2 times. => Number of possible arrangements =IE=5! / 2!Now, the two vowels can be arranged in60= 2 ways. => Total number of possible words such that the vowels are always together =2!,60 x 2 = 120

total number of possible words ==6! / 2!Therefore, the total number of possible words such that the vowels are never together720 / 2 = 360240

## Related Resources:

Test your knowledge of Permutation and Combination in Quantitative Aptitude with the quiz linked below, containing numerous practice questions to help you master the topic:-

<< Practice Permutation and Combination Aptitude Questions >>