Class 9 Advanced Maths Chapter 5 Combinatorics Ex 5.2 Solutions

Students often refer to NCERT Class 9 Advanced Maths Solutions Chapter 5 Combinatorics Ex 5.2 to verify their answers.

Advanced Maths Class 9 Exercise 5.2 Solutions

Class 9 Advanced Maths Ex 5.2 Solutions

Question 1.
Find the HCF and LCM of 6!, 5!.
Solution:
The LCM will be the factorial of the larger number and HCF will be the factorial of the smaller number.
HCF = 5! = 120
LCM = 6! = 720

Question 2.
Find n if \(\frac{n!}{(n-3)!}\) = 60.
Solution:
\(\frac{n!}{(n-3)!}=\frac{n(n-1)(n-2)(n-3)!}{(n-3)!}\) = 60
n(n – 1)(n – 2) = 60 = 5 × 4 × 3
n(n – 1)(n – 2) = 5 (5 – 1)(5 – 2)
n = 5

Question 3.
Find x, if \(\frac{1}{5!}+\frac{1}{6!}=\frac{x}{7!}\)
Solution:
\(\frac{1}{5!}+\frac{1}{6!}=\frac{x}{7!}\)
\(\frac{1}{5!}+\frac{1}{6 \times 5!}=\frac{x}{7 \times 6 \times 5!}\)
1 + \(\frac{1}{6}=\frac{x}{42}\)
x = \(\frac{7}{6}\) × 42 = 49

Question 4.
Find the value(s) of x in each of the following: (Here, x ≥ 0)
(a) x! = 120
Solution:
x! = 120 = 5!
x = 5

(b) (x!)² = 576
Solution:
(x!)² = 576
x! = \(\sqrt{576}\) = 24
x! = 4!
x = 4

(c) (x!)² – 25x! + 24 = 0
Solution:
Let x!= y
Then, y² – 25y + 24 = 0
y² – 24y – y + 24 = 0
(y – 24) (y – 1) = 0
So, y = 24 or y = 1
That means,
x! = 24 ⇒ x = 4
and
x! = 1 ⇒ x = 0 or x = 1
Therefore, x = 0, 1, 4

(d) \(\frac{(x+2)!-(x+1)!}{x!}\) = 49
Solution:
\(\frac{(x+2)!-(x+1)!}{x!}\) = 49
\(\frac{(x+2)(x+1) x!-(x+1) x!}{x!}\) = 49
(x + 2)(x + 1) – (x + 1) = 49
(x + 1)² = 49
x + 1 = 7
x = 6