2nd PUC Basic Maths Question Bank Chapter 2 Permutations and Combinations Ex 2.2

Students can Download Basic Maths Exercise 2.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 2 Permutations and Combinations Ex 2.2

Part – A

2nd PUC Basic Maths Permutations and Combinations Ex 2.2 One Mark Questions and Answers

Question 1.
Find the total number of ways in which 8 different coloured beads can be strung together to form a necklace.
Answer:
(n -1)! = (8 -1)! = 7!.

Question 2.
In how many ways can 9 flowers of different colours be strung together to form a garland.
Answer:
\(\frac{(n-1) !}{2}=\frac{8 !}{2}\)

Question 3.
In how many ways can 10 people be seated around a table.
Answer:
(10 -1)! = 9!

Part B

2nd PUC Basic Maths Permutations and Combinations Ex 2.2 Two Marks Questions and Answers

Question 1.
Find the number of ways in which 8 men be arranged round a table so that 2 particular men may not be next to each other.
Answer:
Total number of ways when there is no restriction is 7!

When 2 particular men are together can be takes as 1 unit and the remaining 6, total = 7 can sit round a table in 6! ways and 2 men can themselves can be done in 2! ways can be done in 6!.2!.
∴ The number of ways in which 2 particular men are not together = 7! – 6! . 2!

Question 2.
In how many ways can 6 gentlemen and 4 ladies be seated round a table so that no two ladies are together.
Answer:
There are 6 vacant places in between the 6 gentlemen and 4 ladies can occupy these gaps in ⁶P₄ ways. For each of these the six gentlemen can be permuted in (6 – 1)! = 5!
∴ The number of ways is 5!⁶P₄

Question 3.
In how many ways can 10 beads of different colours be strung into a necklace if the red, green and yellow beads are always together.
Answer:
Red, yellow and green are together can be permuted in 3! ways and the remaining 7 & 1 unit of other beads altogether 8 beads can permuted in 8! Ways.
∴ The number of ways is 8! 3!.

Question 4.
In how many ways can 6 boys and 6 girls be arranged in a circle so that no two boys are together.
Answer:
Six boys can arranged in a circle in (6 – 1)! = 5!, 6 girls can permute in 6! Ways.
∴ The number of ways is 6! × 5!.

Question 5.
In how many ways can 7 gentlemen and 5 ladies be arranged in a circle if no two ladies are together.
Answer:
Ladies have to be arranged between gentlemen. The 7 gentlemen can be seated in 6! Ways. There are 7 gaps between the men which the 5 ladies can occupy in ⁷P₅, ways.
∴ The number of ways is ⁷P₆.6!.

Question 6.
Find the number of ways in which 10 flowers can be strung into a garland if 3 particular flowers are always together.
Answer:
Three particular flowers are always together can be done in 3! Ways 7 + 1 units of 3 = 8 can be done in 7! Ways.
∴ The total number of ways = 7!. 3!.

Question 7.
Find the number of ways in which 15 staff members can be seated around a circular table for a meeting if the vice-principal and dean have to be an either side of the principal.
Answer:
Vice principal & dean can be permuted in 2 ways. The remaining 13 can be permuted in 12! Ways.
∴ The number of ways is 12! · 2!.