RD Chapter 16 Permutations Ex 16.3 Solutions
Question - 1 : - Evaluate each of the following:
(i) 8P3
(ii) 10P4
(iii) 6P6
(iv) P (6, 4)
Answer - 1 : -
(i) 8P3
We know that, 8P3 canbe written as P (8, 3)
By using the formula,
P (n, r) = n!/(n – r)!
P (8, 3) = 8!/(8 – 3)!
= 8!/5!
= (8 × 7 × 6 × 5!)/5!
= 8 × 7 × 6
= 336
∴ 8P3 =336
(ii) 10P4
We know that, 10P4 canbe written as P (10, 4)
By using the formula,
P (n, r) = n!/(n – r)!
P (10, 4) = 10!/(10 –4)!
= 10!/6!
= (10 × 9 × 8 × 7 ×6!)/6!
= 10 × 9 × 8 × 7
= 5040
∴ 10P4 =5040
(iii) 6P6
We know that, 6P6 canbe written as P (6, 6)
By using the formula,
P (n, r) = n!/(n – r)!
P (6, 6) = 6!/(6 – 6)!
= 6!/0!
= (6 × 5 × 4 × 3 × 2 ×1)/1 [Since, 0! = 1]
= 6 × 5 × 4 × 3 × 2 ×1
= 720
∴ 6P6 =720
(iv) P (6, 4)
By using the formula,
P (n, r) = n!/(n – r)!
P (6, 4) = 6!/(6 – 4)!
= 6!/2!
= (6 × 5 × 4 × 3 ×2!)/2!
= 6 × 5 × 4 × 3
= 360
∴ P (6, 4) = 360
Question - 2 : - If P (5, r) = P (6, r – 1), find r.
Answer - 2 : -
Given:
P (5, r) = P (6, r –1)
By using the formula,
P (n, r) = n!/(n – r)!
P (5, r) = 5!/(5 – r)!
P (6, r-1) = 6!/(6 –(r-1))!
= 6!/(6 – r + 1)!
= 6!/(7 – r)!
So, from the question,
P (5, r) = P (6, r –1)
Substituting theobtained values in above expression we get,
5!/(5 – r)! = 6!/(7 –r)!
Upon evaluating,
(7 – r)! / (5 – r)! =6!/5!
[(7 –r) (7 – r – 1) (7 – r – 2)!] / (5 – r)! = (6 × 5!)/5!
[(7 –r) (6 – r) (5 – r)!] / (5 – r)! = 6
(7 – r) (6 – r) = 6
42 – 6r – 7r + r2 =6
42 – 6 – 13r + r2 =0
r2 –13r + 36 = 0
r2 –9r – 4r + 36 = 0
r(r – 9) – 4(r – 9) =0
(r – 9) (r – 4) = 0
r = 9 or 4
For, P (n, r): r ≤ n
∴ r =4 [for, P (5, r)]
Question - 3 : - If 5 P(4, n) = 6 P(5, n – 1), find n.
Answer - 3 : -
Given:
5 P(4, n) = 6 P(5, n –1)
By using the formula,
P (n, r) = n!/(n – r)!
P (4, n) = 4!/(4 – n)!
P (5, n-1) = 5!/(5 –(n-1))!
= 5!/(5 – n + 1)!
= 5!/(6 – n)!
So, from the question,
5 P(4, n) = 6 P(5, n –1)
Substituting theobtained values in above expression we get,
5 × 4!/(4 – n)! = 6 ×5!/(6 – n)!
Upon evaluating,
(6 – n)! / (4 – n)! =6/5 × 5!/4!
[(6 –n) (6 – n – 1) (6 – n – 2)!] / (4 – n)! = (6 × 5 × 4!) / (5 × 4!)
[(6 –n) (5 – n) (4 – n)!] / (4 – n)! = 6
(6 – n) (5 – n) = 6
30 – 6n – 5n + n2 =6
30 – 6 – 11n + n2 =0
n2 – 11n+ 24 = 0
n2 –8n – 3n + 24 = 0
n(n – 8) – 3(n – 8) =0
(n – 8) (n – 3) = 0
n = 8 or 3
For, P (n, r): r ≤ n
∴ n =3 [for, P (4, n)]
Question - 4 : - If P(n, 5) = 20 P(n, 3), find n.
Answer - 4 : -
Given:
P(n, 5) = 20 P(n, 3)
By using the formula,
P (n, r) = n!/(n – r)!
P (n, 5) = n!/(n – 5)!
P (n, 3) = n!/(n – 3)!
So, from the question,
P(n, 5) = 20 P(n, 3)
Substituting theobtained values in above expression we get,
n!/(n – 5)! = 20 ×n!/(n – 3)!
Upon evaluating,
n! (n – 3)! / n! (n –5)! = 20
[(n –3) (n – 3 – 1) (n – 3 – 2)!] / (n – 5)! = 20
[(n –3) (n – 4) (n – 5)!] / (n – 5)! = 20
(n – 3) (n – 4) = 20
n2 –3n – 4n + 12 = 20
n2 –7n + 12 – 20 = 0
n2 –7n – 8 = 0
n2 –8n + n – 8 = 0
n(n – 8) – 1(n – 8) =0
(n – 8) (n – 1) = 0
n = 8 or 1
For, P(n, r): n ≥ r
∴ n =8 [for, P(n, 5)]
Question - 5 : - If nP4 = 360, find the value of n.
Answer - 5 : -
Given:
nP4 = 360
nP4 can be written as P (n , 4)
By using the formula,
P (n, r) = n!/(n – r)!
P (n, 4) = n!/(n – 4)!
So, from the question,
nP4 = P (n, 4) = 360
Substituting theobtained values in above expression we get,
n!/(n – 4)! = 360
[n(n– 1) (n – 2) (n – 3) (n – 4)!] / (n – 4)! = 360
n (n – 1) (n – 2)(n – 3) = 360
n (n – 1) (n – 2)(n – 3) = 6×5×4×3
On comparing,
The value of n is 6.
Question - 6 : - If P(9, r) = 3024, find r.
Answer - 6 : -
Given:
P (9, r) = 3024
By using the formula,
P (n, r) = n!/(n – r)!
P (9, r) = 9!/(9 – r)!
So, from the question,
P (9, r) = 3024
Substituting theobtained values in above expression we get,
9!/(9 – r)! = 3024
1/(9 – r)! = 3024/9!
= 3024/(9×8×7×6×5×4×3×2×1)
=3024/(3024×5×4×3×2×1)
= 1/5!
(9 – r)! = 5!
9 – r = 5
-r = 5 – 9
-r = -4
∴ The value of ris 4.
Question - 7 : - If P (11, r) = P (12, r – 1), find r.
Answer - 7 : -
Given:
P (11, r) = P (12, r –1)
By using the formula,
P (n, r) = n!/(n – r)!
P (11, r) = 11!/(11 –r)!
P (12, r-1) = 12!/(12– (r-1))!
= 12!/(12 – r + 1)!
= 12!/(13 – r)!
So, from the question,
P (11, r) = P (12, r –1)
Substituting theobtained values in above expression we get,
11!/(11 – r)! =12!/(13 – r)!
Upon evaluating,
(13 – r)! / (11 – r)!= 12!/11!
[(13– r) (13 – r – 1) (13 – r – 2)!] / (11 – r)! = (12×11!)/11!
[(13– r) (12 – r) (11 -r)!] / (11 – r)! = 12
(13 – r) (12 – r) = 12
156 – 12r – 13r + r2 =12
156 – 12 – 25r + r2 =0
r2 –25r + 144 = 0
r2 –16r – 9r + 144 = 0
r(r – 16) – 9(r – 16)= 0
(r – 9) (r – 16) = 0
r = 9 or 16
For, P (n, r): r ≤ n
∴ r =9 [for, P (11, r)]
Question - 8 : - If P(n, 4) = 12. P(n, 2), find n.
Answer - 8 : -
Given:
P (n, 4) = 12. P (n,2)
By using the formula,
P (n, r) = n!/(n – r)!
P (n, 4) = n!/(n – 4)!
P (n, 2) = n!/(n – 2)!
So, from the question,
P (n, 4) = 12. P (n,2)
Substituting theobtained values in above expression we get,
n!/(n – 4)! = 12 ×n!/(n – 2)!
Upon evaluating,
n! (n – 2)! / n! (n –4)! = 12
[(n –2) (n – 2 -1) (n – 2 – 2)!] / (n – 4)! = 12
[(n –2) (n – 3) (n – 4)!] / (n – 4)! = 12
(n – 2) (n – 3) = 12
n2 –3n – 2n + 6 = 12
n2 –5n + 6 – 12 = 0
n2 –5n – 6 = 0
n2 –6n + n – 6 = 0
n (n – 6) – 1(n – 6) =0
(n – 6) (n – 1) = 0
n = 6 or 1
For, P (n, r): n ≥ r
∴ n =6 [for, P (n, 4)]
Question - 9 : - If P(n – 1, 3) : P(n, 4) = 1 : 9, find n.
Answer - 9 : -
Given:
P (n – 1, 3): P (n, 4)= 1 : 9
P (n – 1, 3)/ P (n, 4)= 1 / 9
By using the formula,
P (n, r) = n!/(n – r)!
P (n – 1, 3) = (n –1)! / (n – 1 – 3)!
= (n – 1)! / (n – 4)!
P (n, 4) = n!/(n – 4)!
So, from the question,
P (n – 1, 3)/ P (n, 4)= 1 / 9
Substituting theobtained values in above expression we get,
[(n –1)! / (n – 4)!] / [n!/(n – 4)!] = 1/9
[(n –1)! / (n – 4)!] × [(n – 4)! / n!] = 1/9
(n – 1)!/n! = 1/9
(n – 1)!/n (n – 1)! =1/9
1/n = 1/9
n = 9
∴ The value of nis 9.
Question - 10 : - If P(2n – 1, n) : P(2n + 1, n – 1) = 22 : 7 find n.
Answer - 10 : -
Given:
P(2n – 1, n) : P(2n +1, n – 1) = 22 : 7
P(2n – 1, n) / P(2n +1, n – 1) = 22 / 7
By using the formula,
P (n, r) = n!/(n – r)!
P (2n – 1, n) = (2n –1)! / (2n – 1 – n)!
= (2n – 1)! / (n – 1)!
P (2n + 1, n – 1) =(2n + 1)! / (2n + 1 – n + 1)!
= (2n + 1)! / (n + 2)!
So, from the question,
P(2n – 1, n) / P(2n +1, n – 1) = 22 / 7
Substituting theobtained values in above expression we get,
[(2n– 1)! / (n – 1)!] / [(2n + 1)! / (n + 2)!] = 22/7
[(2n– 1)! / (n – 1)!] × [(n + 2)! / (2n + 1)!] = 22/7
[(2n– 1)! / (n – 1)!] × [(n + 2) (n + 2 – 1) (n + 2 – 2) (n + 2 – 3)!] / [(2n + 1)(2n + 1 – 1) (2n + 1 – 2)] = 22/7
[(2n– 1)! / (n – 1)!] × [(n + 2) (n + 1) n(n – 1)!] / [(2n + 1) 2n (2n – 1)!] =22/7
[(n +2) (n + 1)] / (2n + 1)2 = 22/7
7(n + 2) (n + 1) =22×2 (2n + 1)
7(n2 +n + 2n + 2) = 88n + 44
7(n2 +3n + 2) = 88n + 44
7n2 +21n + 14 = 88n + 44
7n2 +21n – 88n + 14 – 44 = 0
7n2 –67n – 30 = 0
7n2 –70n + 3n – 30 = 0
7n(n – 10) + 3(n – 10)= 0
(n – 10) (7n + 3) = 0
n = 10, -3/7
We know that, n ≠ -3/7
∴ The value of nis 10.