Permutation and Combination Practice Questions

Practice Questions on Permutation and Combination

In this page you will find out Permutation and Combination Questions and Answers  and Formulas to solve the Permutations and Combinations based Questions. Go through this page to clear whole concepts of Permutation and Combination.
Further on this page, you will get sample questions of Permutation and Combination.
Practice Questions for Permutation and Combination

About Permutation & Combination

Points to remember

  • There are certain rules that are specified for permutation and combinations. These rules contain formulas and basic know-how which need to be considered while solving questions related to the same. It will help you to understand the concept in a better way and to avoid mistakes in the future. To through the below given rules and formulas for gaining further understanding on the subject:
  • The number of all combinations of n things, taken r at a time is:
    ^{n}C_{r} = n! = n (n – 1) (n – 2) … to r factors.
    \frac{(n!)}{(n – r)! r!}
  • While solving questions on combinations, you should know the below-mentioned formulas:
    o ^{n}C_{n} = 1 \text{ and } ^{n}C_{0} = 1
    o ^{n}C_{r} = ^{n}C_{(n – r)}
  • The permutation of n events, taken r at a time, is represented as:
    ^{n}P_{r} = n(n – 1)(n – 2) … (n – r + 1) = \frac{n!}{(n – r)!}
  • Key point to know is that the statement questions are chiefly made to check your knowledge to make an equation and then put formulae for it.
  • Permutations are used when an issue calls for the number of activities of things as well as various tips are to be calculated.
  • Combinations are used when an issue arise for the number of tricks of choosing things along with the order of choice is not to be calculated.

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Permutation and Combination Questions and Answers

1. Among 9 consonants and 5 vowels, what number of words can be formed of 4 consonants and 3 vowels?

6400400

6400400

25%

6350400

6350400

56.82%

6550500

6550500

9.09%

6350600

6350600

9.09%

The number of ways to choose 4 consonants from 9 is given by the combination formula:

Number of ways to choose 4 consonants = 9C4 = 9! / (4! * (9-4)!) = 9! / (4! * 5!) = (9 * 8 * 7 * 6) / (4 * 3 * 2 * 1) = 126

Similarly, the number of ways to choose 3 vowels from 5 is given by:

Number of ways to choose 3 vowels = 5C3 = 5! / (3! * (5-3)!) = 5! / (3! * 2!) = (5 * 4) / (2 * 1) = 10

Now, we can combine the chosen consonants and vowels to form words. Since the order matters, we can use the multiplication principle.

Number of words = (Number of ways to choose 4 consonants) * (Number of ways to choose 3 vowels) = 126 * 10 = 1260

Therefore, the number of words that can be formed with 4 consonants and 3 vowels is 1260.

the 7 letters can be arranged in 7! ways :

= 1260 * 7!
=  1260 *5040

2. There are 4 bananas, 7 apples and 6 mangoes present in the fruit basket of kareena kapoor who went to supermarket . In how many ways can a person make a selection of fruits from the basket.

269

269

11.11%

256

256

33.33%

279

279

38.89%

280

280

16.67%

Zero or more bananas can be selected in 4 + 1 = 5 ways (0 orange, 1 orange, 2 orange, 3 orange and 4 orange)
similarly apples can be selected in 7 +1 = 8 ways
and mangoes in 6 +1 = 7 ways
so total number of ways = 5*8*7 = 280
but we included a case of 0 orange, 0 apple and 0 mangoes, so we have to subtract this, so 280 – 1 = 279 ways

3. Find out the number of arrangements that can be made for the word RAINBOW if all the the vowels are placed on the even places?

720

720

25.58%

144

144

46.51%

120

120

13.95%

36

36

13.95%

The word RAINBOW has 7 letters in which vowels are 3.

As mentioned in the question the vowels have to reside in even places, so they can be set in the 3 even positions in 3! So 3x2x1=6 ways. Also the consonants can be set in the left over 4 positions in 4! So 4x3x2x1=24 ways.

Therefore the total number of ways will be: 24 * 6 = 144.

4. In how many ways the word 'MESMERISE' can be written?

5!/(2!)2 3!

5!/(2!)2 3!

22.73%

9!/(2!)3 3!

9!/(2!)3 3!

31.82%

9!/(2!)2 (2!)3

9!/(2!)2 (2!)3

38.64%

10!/(2!*3!)

10!/(2!*3!)

6.82%

Clearly the word 'MESMERISE' contains of 10 letters in which there are M are 2 E are 3, S are 2 and I and R are 1.

So the total number of arrangements are= 10!/(2!*3!.

5. A panel has total of 11 members 5 males and 6 females. Find out the number of ways of picking 2 males and 3 females from the given panel team?

110

110

7.89%

300

300

13.16%

200

200

68.42%

350

350

10.53%

The total number of ways to pick two male and three female = ⁵C₂ * ⁶C₃

= 200

6. There are 21 stations between Chennai and Darjeeling. Find out that what is the number of second class tickets have to be produced, so that a traveler can travel from any point of station to the any of the station?

506

506

34.38%

380

380

40.63%

950

950

9.38%

180

180

15.63%

The total number of stations = 23

From 23 stations we need to select any 2 stations and the direction of travel (i.e., Chennai to Darjeeling is different from Darjeeling to Chennai) in ²³P₂ ways.

²³P₂ = 23 * 22 = 506.

7. A device password comprises 5 rings each having 7 different messages. Find out the maximum number of different unsuccessful efforts to open the lock?

22316

22316

16.13%

24321

24321

22.58%

16806

16806

51.61%

17283

17283

9.68%

As each ring have 7 letters. Therefore, the number of tries made with the five rings is = 7 * 7 * 7 * 7 * 7 = 16807. Out of these efforts, one will be successful effort.

Maximum number of unsuccessful efforts = 16807 - 1 = 16806.

8. Find out the number of ways in which 6 male teachers and 6 female teachers can be settled on seat for a shoot so that no two female teachers sit simultaneously?

2 (6!)

2 (6!)

37.84%

6! * ⁷P₆

6! * ⁷P₆

37.84%

1(6!) 2

1(6!) 2

10.81%

9! * 6

9! * 6

13.51%

As given first we will arrange all six male teachers in 6! Ways.

Therefore, now 6 female teachers require to be arranged on 7 seats and it can be done in ⁷P₆ ways.

Thus, number of ways = 6! * ⁷P₆

9. By using the digits (1, 3, 4, 5, 6, 7, 8) how many five digit numbers can be made? Rule*(Recurrence of numbers is not allowed)?

3610

3610

12.5%

6020

6020

15.63%

2520

2520

68.75%

2180

2180

3.13%

We have total of seven digits.

Five digit numbers which can be formed in ways = ⁷P⁵

= 7* 6 * 5 * 4 * 3 = 2520

10. Out of numbers (1, 2,3,5,7 & 9) how many four digit even numbers can be formed?

100

100

9.09%

360

360

30.3%

60

60

48.48%

260

260

12.12%

The numbers we have: 1, 2, 3, 5, 7, 9

Out of the given numbers, there is only one even digit number i.e.2. So elements place is occupied with only '2' and the left over 3 positions can be filled as: ⁵P₃ ways.

So the number of even numbers = ⁵P₃ = 60.

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