Class 9 Advanced Maths Chapter 5 Combinatorics Ex 5.3 Solutions

Students often refer to NCERT Class 9 Advanced Maths Solutions Chapter 5 Combinatorics Ex 5.3 to verify their answers.

Advanced Maths Class 9 Exercise 5.3 Solutions

Class 9 Advanced Maths Ex 5.3 Solutions

Question 1.
In how many ways can 4 distinct cars be parked in 6 empty spaces?
Solution:
The number of ways of parking four distinct cars in six empty spaces = \({ }^6 P_4=\frac{6!}{(6-4)!}=\frac{6!}{2!}\)
= 6 × 5 × 4 × 3 = 360

Question 2.
How many 3-letter words (with or without meaning) can be formed using the letters of the word “LOGIC”?
Solution:
Word: LOGIC There are 5 distinct letters.
3-letter words can be formed by arranging 3 letters from 5. ⁵P₃ = 5 × 4 × 3 = 60 words

Question 3.
A library has 5 distinct science books and 3 distinct math books. In how many ways can they be arranged on a shelf if the math books must occupy the first 3 positions?
Solution:
In the first 3 positions, the math books can be arranged in ³P₃ ways and the science books can be arranged in ⁵P₅ ways.
Required number of ways
= ³P₃ × ⁵P₅ = 3! × 5!
= 6 × 120 = 720

Question 4.
In how many ways can 5 boys and 2 girls be seated in a row of 7 chairs if the 2 girls must always sit together?
Solution:
Since the two girls must always sit together, we can think of them as forming a single block. Now we have, 5 boys and 1 block of girls. So, we have to arrange total 6 units.
Now, the 6 units can be arranged in a row in 6! ways i.e., 720 ways.

The girls can be arranged among themselves in 2! ways. By the Fundamental Principle of Multiplication, the total number of ways are 720 × 2 = 1440 ways.

Question 5.
How many 4-digit numbers can be formed using the digits 2, 4, 6, 8, 9 (without repetition) if the number must be strictly greater than 6000?
Solution:
Digits: 2, 4, 6, 8, 9
Number must be 4-digit and greater than 6000.
Repetition is not allowed.
Thousands place can be 6, 8, or 9.
Case 1: Thousands place = 6
Remaining 3 places from 4 digits:
4 × 3 × 2 = 24
Case 2: Thousands place = 8
4 × 3 × 2 = 24
Case 3: Thousands place = 9
4 × 3 × 2 = 24
Total Numbers: 24 + 24 + 24 = 72

Question 6.
How many words (with or without meaning) can be formed from the letters of the word, ‘DAUGHTER’, so that:
(i) all vowels occur together?
(ii) all vowels do not occur together?
Solution:
(i) Treat the 3 vowels as a single block. So instead of 8 letters, we now have:
(Vowel block) + 5 consonants = 6 units
These 6 units can be arranged in 6! = 720 ways.
The vowels among themselves can be arranged in 3! = 6 ways
Number of words with vowels together = 4320

(ii) Total arrangements of all 8 distinct letters = 8! = 40320
Number of words with vowels together = 4320
Number of words with all vowels not together
= 40320 – 4320 = 36000