Permutation and Combination Aptitude

Permutation and Combination are fundamental concepts in probability and counting theory, used in determining the number of ways objects can be arranged (permutation) and chosen (combination). Permutations focus on the arrangement of items where the order is significant, while combinations deal with the selection of items where the order is not important.

1. How many different 3-letter words can be formed using the letters A, B, C, D, and E without repeating any letter?

Number of ways = 5P3 = 5!/(5-3)! = 5 × 4 × 3 = 60.

2. In how many ways can a committee of 3 be formed from a group of 4 women and 6 men?

Number of ways = 10C3 = 10!/(3! × 7!) = 120.

3. How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated?

Number of ways = 5P4 = 5!/(5-4)! = 5 × 4 × 3 × 2 = 120.

4. If 5 different books are to be arranged on a shelf, how many different arrangements are possible?

Number of ways = 5! = 5 × 4 × 3 × 2 × 1 = 120.

5. From a group of 7 men and 5 women, how many different teams of 5 people can be formed that include exactly 3 women?

Number of ways = 5C3 × 7C2 = (5!/(3! × 2!)) × (7!/(2! × 5!)) = 350.

6. How many different 2-letter words can be formed from the letters A, B, C, D, and E if repetition of letters is allowed?

Number of ways = 5 × 5 = 25 (since repetition is allowed).

7. In how many ways can 6 people be seated in a round table?

Number of ways = (6-1)! = 5! = 5 × 4 × 3 × 2 × 1 = 120.

8. How many ways can a team of 3 men and 2 women be selected from 5 men and 4 women?

Number of ways = 5C3 × 4C2 = (5!/(3! × 2!)) × (4!/(2! × 2!)) = 60.

9. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4, 5 if repetition of digits is not allowed?

Number of ways = 5P3 = 5!/(5-3)! = 5 × 4 × 3 = 60.

10. How many different 4-letter arrangements can be formed from the word 'LEAD'?

Number of ways = 4! = 4 × 3 × 2 × 1 = 24.

11. In how many ways can 5 books on different subjects be arranged on a shelf?

Number of ways = 5! = 5 × 4 × 3 × 2 × 1 = 120.

12. How many 4-digit numbers can be formed using the digits 1, 2, 3, and 4, without repetition?

Number of ways = 4P4 = 4! = 4 × 3 × 2 × 1 = 24.

13. How many triangles can be formed by joining 8 points on a plane, no three of which are collinear?

Number of ways = 8C3 = 8!/(3! × 5!) = 56.

14. From a group of 6 ladies and 4 gentlemen, a committee of 3 members is to be formed. In how many ways can this be done if the committee must have at least one lady?

Total ways = 10C3 - 4C3 (excluding all men groups) = 120 - 4 = 116.

15. A password consists of 4 distinct alphabets. How many such passwords can be formed from the alphabets A, B, C, D, E, F?

Number of ways = 6P4 = 6!/(6-4)! = 6 × 5 × 4 × 3 = 360.

16. How many ways can the letters of the word 'ORANGE' be arranged so that the vowels always come together?

Treat the vowels OAE as one unit. Then we have 4 units (OAE, R, N, G). Arrange these 4 units in 4! ways and arrange the vowels OAE among themselves in 3! ways. Total = 4! × 3! = 24 × 6 = 144.

17. In how many ways can 7 friends be seated in a row if two of them insist on sitting next to each other?

Treat the two friends who want to sit together as one unit. Now there are 6 units (5 individual friends and 1 pair). Arrange these 6 units in 6! ways and the pair among themselves in 2! ways. Total = 6! × 2! = 720 × 2 = 1440.

18. How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6, 7 if repetition of digits is allowed?

Number of ways = 7 × 7 × 7 = 7^3 = 343.

19. A committee of 4 is to be selected from 6 men and 5 women. In how many ways can this be done if the committee must have at least 2 women?

Total ways = 5C2 × 6C2 + 5C3 × 6C1 + 5C4 = 150 + 20 + 5 = 175.

20. How many ways can 5 boys and 5 girls form a circle such that all boys stand together?

Treat all boys as one unit. Now there are 6 units (5 girls and 1 group of boys). Arrange these 6 units in a circle in (6-1)! ways. Arrange the boys among themselves in 5! ways. Total = 5! × 5! = 120 × 120 = 14400.