Practicing the AP Board Solutions Class 11 Maths and Chapter 6 Inter 1st Year Maths Permutations and Combinations Solutions Exercise 6b Pdf Download will help students to clear their doubts quickly.
Intermediate 1st Year Maths Permutations and Combinations Solutions Exercise 6b
Permutations and Combinations Exercise 6b Solutions
Permutations and Combinations Class 11 Exercise 6b Solutions – Permutations and Combinations 6b Exercise Solutions
Question 1.
Evaluate
(i) 8!
(ii) 4! – 3!
Solution:
(i) Consider 8!
We know that
8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320
(ii) Consider 4! – 3!
4! – 3! = (4 × 3!) – 3!
The above equation can be written as 3! (4 – 1)
= 3 × 2 × 1 × 3
= 18
Question 2.
Is 3! + 4! = 7!?
Solution:
Consider L.H.S = 3! + 4!
Computing the left-hand side, we get
31 + 4! = (3 × 2 × 1) + (4 × 3 × 2 × 1)
= 6 + 24
= 30
Again, considering RHS and computing, we get
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
Therefore, 3! + 4! ≠ 7!
Question 3.
Compute \(\frac{8!}{6!\times 2!}\)
Solution:
Given \(\frac{8!}{6!\times 2!}\)
Expanding all the factorials and simplifying, we get
\(\frac{8!}{6!\times 2!}=\frac{8 \times 7 \times 6!}{6!\times 2 \times 1}\)
= \(\frac{8 \times 7}{2}\)
= 28
Question 4.
If \(\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}\), find x.
Solution:
Consider L.H.S. and by computing
we get \(\frac{1}{6!}+\frac{1}{7!}=\frac{1}{6!}+\frac{1}{7 \times 6!}\)
⇒ \(\frac{7+1}{7 \times 6!}=\frac{8}{7!}\)
Equating L.H.S. to R.H.S. to get the value of x.
\(\frac{8}{7!}=\frac{x}{8!}\)
⇒ \(\frac{8}{7!}=\frac{x}{8 \times 7!}\)
On rearranging, we get
8 × 8 = x
⇒ x = 64
Question 5.
Evaluate \(\frac{n!}{(n-r)!}\), when
(i) n = 6, r = 2
(ii) n = 9, r = 5
Solution:
(i) Given n = 6 and r = 2
Putting the value of n and r we get
\(\frac{6!}{(6-2)!}\) = \(\frac{6!}{4!}\)
= \(\frac{6 \times 5 \times 4!}{4!}\)
= 6 × 5
= 30
(ii) Given n = 9 and r = 5
Putting the value of n and r, we get
\(\frac{9!}{(9-5)!}\) = \(\frac{9!}{4!}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!}\)
= 9 × 8 × 7 × 6 × 5
= 15120