Referring to the AP Inter 1st Year Maths Study Material Chapter 5 Permutations and Combinations Exercise 6b Solutions makes it easier to understand complex problems.
AP Inter 1st Year Maths Permutations and Combinations Solutions Exercise 6b
Question 1.
Evaluate (i) 8! (ii) 4! – 3!
Solution:
i) 8! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40320
ii) 4! = 1 × 2 × 3 × 4 = 24 ⇒ 3! = 1 × 2 × 3 = 6
∴ 4! – 3! = 24 – 6 = 8
Question 2.
Is 3! + 4! = 7! ?
Solution:
3! = 1 × 2 × 3 = 6 and 4! = 1 × 2 × 3 × 4 = 24 ⇒ 3! + 4! = 6 + 24 = 30
But 7! = 1 × 2 × 3 × 4 × 5 × 6 × 7 = 5040; Hence, 3! + 4! ≠ 71.
Question 3.
Compute \(\frac{8!}{6!2!}\)
Solution:
\(\frac{8!}{6!2!}\)
= \(\frac{8 \times 7}{2}\)
= 28
Question 4.
If \(\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}\) find x.
Solution:
⇒ \(\frac{1}{6!} \times \frac{7}{7}+\frac{1}{7!}=\frac{x}{8!}\)
⇒ \(\frac{7}{7!}+\frac{1}{7!}=\frac{x}{8!}\)
⇒ \(\frac{8}{7!}\) × 8! = x
⇒ \(\frac{8}{7!}\) × 8 × 7! = x
⇒ x = 64
Question 5.
Evaluate \(\frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{r})!}\), when
i) n = 6, r = 2
Solution:
n = 6, r = 2
\(\frac{n!}{(n-r)!}=\frac{6!}{(6-2)!}=\frac{6!}{4!}\)
= 30
ii) n = 9, r = 5
Solution:
n = 9, r = 5
\(\frac{n!}{(n-r)!}=\frac{9!}{(9-5)!}=\frac{9!}{4!}\)
= 9 × 8 × 7 × 6 × 5
= 15120