Chapter 16: Permutations – Exercise 16.1
Permutations – Exercise – 16.1 – Q.1
We have,
= 30 × 29
= 870
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(ii) We have
.png "Permutations – Exercise – 16.1 – Q.1(ii)")
= 10 × 10
= 100
(iii) We have,
8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
and 6! = 6 × 5 × 4 × 3 × 2 × 1
∴ L. C. M. (6!, 7!, 8!) = 8!
Permutations – Exercise – 16.1 – Q.2
L.H.S:
Permutations – Exercise – 16.1 – Q.3(i)
We have,
.png "Permutations – Exercise – 16.1 – Q.3(i)")
⟹ x = 6 × 6
⟹ x = 36.
Hence, x = 36.
Permutations – Exercise – 16.1 – Q.3(ii)
.png "Permutations – Exercise – 16.1 – Q.3(ii)")
⟹ x = 10 × 9 + 10
⟹ x = 100
Permutations – Exercise – 16.1 – Q.3(iii)
.png "Permutations – Exercise – 16.1 – Q.3(iii)")
⟹ x = 8 × 7 + 8
⟹ x = 64
Permutations – Exercise – 16.1 – Q.4(i)
We have,
5 × 6 × 7 × 8 × 9 × 10
.png "Permutations – Exercise – 16.1 – Q.4(i)")
Permutations – Exercise – 16.1 – Q.4(ii)
We have,
3 × 6 × 9 × 12 × 15 × 18
= 3 ×(3 × 2) × (3 × 3) × (3 × 4) × (3 × 5) × (3 × 6)
= 36 × [2 × 3 × 4 × 5 × 6]
= 36 × (6!)
Permutations – Exercise – 16.1 – Q.4(iii)
We have,
(n + 1)(n + 2)(n + 3)…………(2n)
= [1 × 2 × 3 × 4………..(n - 1)n] × (n + 1)(n + 2) .. (2n - 1) × 2n
[1 × 2 × 3 × 4………(n - 1)n]
= (2n!)!/n!
Permutations – Exercise – 16.1 – Q.4(iv)
We have,
1 × 5 × 7 × 9………….. (2n - 1)
.png "Permutations – Exercise – 16.1 – Q.4(iv)")
Permutations – Exercise – 16.1 – Q.5
(i) LHS = (2 + 3)!
= 5!
= 5 × 4 × 3 × 2 × 1
= 120
and, RHS = 2! + 3!
= 2 × 1 + 3 × 2
= 2 × 1 + 3 × 2 × 1
= 2 + 6
= 8
∴ 120 ≠ 8
∴ (2 + 3)! ≠ 2! + 3!
So, it is false.
(ii) LHS = (2 × 3)!
= 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720
and, RHS = 2! × 3!
∴ 720 ≠ 12
∴ (2 × 3)! ≠ 2! × 3!
Hence, it is false.
Permutations – Exercise – 16.1 – Q.6
LHS = n! + (n + 1)!
= n! + (n + 1)(n + 1 - 1)!
= n! + (n + 1)n!
= n!(1 + n + 1)
= n!(n + 2)
= LHS
∴ n!(n + 2) = n! + (n + 1)!
Hence, proved
Permutations – Exercise – 16.1 – Q.7
We have,
(n + 2)! = 60[(n - 1)!]
(n + 2)(n + 1)(n)(n - 1)! = 60[(n - 1)!]
⟹ (n + 2)(n + 1)n = 60
⟹ (n + 2)(n + 1)n = 5 × 4 × 3
∴ n = 3 [By comparing]
Hence, n = 3
Permutations – Exercise – 16.1 – Q.8
We have,
(n + 1)! = 90[(n - 1)!]
⟹ (n + 1) × n × (n - 1)! = 90[(n - 1)!]
⟹ n(n + 1) = 90
⟹ n2 + n = 90
⟹ n2 + n - 90 = 0
⟹ n2 + n - 90 = 0
⟹ n2 + n - 90 = 0
⟹ n2 + 10n - 9n - 90 = 0
⟹ n(n + 10)(n + 10) = 0
⟹ n - 9 = 0 [∵ n + 10 ≠ 0]
⟹ n = 9
Hence, n = 9