UNIT 1: ALGEBRA

Permutation and Combination Class 12 Basic Maths

Exercise 1.2

1. Find the number of permutations of five different object taken three at a time.

Soln: The permutation of n = 5 objects taken r = 3 objects can be made by

P(n, r) = P(5, 3)

= 5! (5–3)!

= 5! 2!

= 5 × 4× 3 = 60 ways.

2. If three persons enter a bus in which there are ten vacant seats, find in how many ways they can sit.

Soln: Three passenger can be arrange in 10 seat by

P(n, r) = P (10, 3)

= 10! (10–3)!

= 10! 7! = 10×9×8×7! 7! = 720 ways

3. a) How may plates of vehicles consisting of 4 different digits can be made out of the integers 4, 5, 6, 7, 8, 9?How many of these numbers are divisible by 2?

Soln: From the 6 integers 4 different digit numbers can be formed by

P(n, r) = P(6, 4) = 6! (6–4)! = 6! 2!

= 6×5×4×3×2×1 2×1 = 360

For divisible by 2; Since the even numbers of 4 digit numbers are divisible by 2. So, the unit place of the number can arranged in 3 ways.Then, the remaining 3 digits of the number can be arranged from 5 integers in P(5, 3) ways.

Now, by the basic principle of counting,

The required number of 4 digits divisible by 2 = 3 × P(5, 3) = 3 × 5! (5 – 3)! = 180

b) How many numbers of 4 different digits can be formed from the digits 2, 3, 4, 5, 6, 7? How many of these numbers are; i) divisible by 5 ii) not divisible by 5

. Soln: Here, total no. of digits 2, 3, 4, 5, 6, 7, n = 6, no. of taken digits for a number, r = 4

Now, by the definition of permutation, The required number of permutations, P(n, r) = P(6, 4)

= 6! (6–4)! = 6×5×4×3×2×1 2×1 = 360

i) For divisible by 5, once place is 5 in the given digits that can be arranged in 1 way. Other 3 digits can be arranged in P(5, 3) ways.

Now, by the basic principle of counting, required number of 4 digits divisible by 5 = 1 × P(5, 3) = 5! (5 – 3)! = 5×4×3×2! 2! = 60

ii) The number can be arranged not divisible by 5 = 360 – 60 = 300

c) How many 5 digit odd numbers can be formed using the digits 3, 4, 5, 6, 7, 8 and 9 if; i) repetition of digits is not allowed ii) repetition of digits is allowed?

Soln: i) To form 5 digit odd number form the digits 3, 4, 5, 6, 7, 8 and 9when repetition not allowed, The arrangement ofunit place has 4 choices (3 or 5 or 7 or 9) and remaining 4 places can be arranged in P(6, 4) = 6! (6–4)! = 6! 2! = 360 ways.

The required number of arrangement if repetition is not allowed = 4  360 = 1440 ways.

ii) To form 5 digit odd number form the digits 3, 4, 5, 6, 7, 8 and 9 and repetition allowed,

The arrangement ofunit place has 4 choices (3 or 5 or 7 or 9) and remaining each 4 places can be arranged in 7 ways.

The number of arrangements if repetition is allowed = 4  7  7  7  7 = 9604 ways.

4. In how many ways can four boys and three girls be seated in a row containing seven seats?

a) if they may sit anywhere

b) if the boys and girls must alternate

c) if all three girls are together?

d) if girls are to occupy odd sets.

Soln: There are 4 boys and 3 girls. Then total no. of persons (n) = 4 + 3 = 7.

Total no. of seats in row (r) = 7

a) The number of arrangements of 7 person in a row = P (7, 7) = 7! (7 – 7)! = 7! 0! = 5040 ways.

b) To arrange the boys and girls alternately 4 boys can be arrange in 4 seatsby P (4, 4) = 4! = 4×3× 2 × 1 = 24 ways 3 girls can be arrange in 4 seatsbyP (3, 3) = 3! = 3×2 × 1 = 6 ways.

 Required ways = 24 × 6 = 144

c) As all three girls are together, let's consider them as single unit, the number of ways that 3 girls together and 4 boys be seated in a row = P (5, 5) = 5! = 5 × 4 × 3 × 2 ×1 = 120 ways

Also, three girls sitting together can sit in p (3, 3) = 3! = 3 × 2 ×1 = 6 ways

 Required ways = 120  6 = 720

d) If girls are to occupy odd sets, the 3 girls can be arrange 4 odd seats by P(4, 3) = 4! = 24 ways

Again, remaining 4 boys can be arrange in remaining 4 seats by P(4, 4) = 4! = 24 ways

Required arrangements = 24  24 = 576 ways.

5. In how many ways can eight people be seated in a row of eight seats so that two particular persons are;

a) always together b) never together?

Soln: Here, the arrangements of 8 people in a row is 8! = 40320 ways

a) When two particular persons are always together, we have to consider these two people as one.

Now, no. arrangements of 7 people on 7 seats = P(7, 7) = 7! = 5040 ways T

he two persons kept together has their P (2, 2)= 2! arrangements 

The required arrangements = P (2, 2) × P (7, 7) = 2! × 7! = 10080 ways.

b) When two particular persons are never together, 

The required arrangements =40320 – 10080 = 30240 ways.

6. Six different books are arranged on a shelf. Find the number of different ways in which two particular books are; a) always together b) not together.

Soln: a) Let's suppose two books as single unit.

Then, no of ways in which two particular books are always together = p (5, 5) = 5!

The two books kept together has their P (2, 2)= 2! arrangements 

Number of required arrangement 5! × 2! = 240 ways.

b) Total no. of arrangement of books = P (6, 6) = 6! = 720 ways.

No. of arrangement in which two particular books are not together = 720 – 240 = 480 ways.

7. In how many ways can four red beads, five white beads and three blue beads be arranged in a row?

Soln: Total no. of beads = 12, in which red beads = 4, white beads = 5 and blue beads = 3.  Required no. of permutations = 12! / 4! 5! 3! = 27720

8. In how many ways can the letters of the following words be arranged?

a) ELEMENT b) NOTATION c) MATHEMATICS d) MISSISSIPPI

Soln:

a) Total no. of letters in the word “ELEMENT” is 7 no. of repetition of E = 3

 Required no of arrangements = 7! 3! =7 × 6 × 5 × 4 × 3! 3! = 840

b) Total no. of letters in the word “NOTATION” is 8

no. of repetition of N= 2

no. of repetition of O = 2 no. of repetition of T= 2

 Required no. of arrangements = 8! 2! 2! 2! = 8  7 × 6 × 5 × 4 × 3  2! 2  2  2! = 5040

c) Total no. of letters in the word "MATHEMATICS" is 11

no. of repetition of M= 2

no. of repetition of A = 2