Each of the arrangements, which can be made by taking some or all of a number of things, is termed a permutation. This exercise mainly deals with permutations and problems based on it. Examples are solved before the exercise-wise problems to help increase knowledge about the fundamental concepts among students. Students can refer to the solutions PDF while solving the problems from the RD Sharma textbook. Students can improve their problem-solving and logical thinking abilities by using RD Sharma Class 11 Maths Solutions PDF, from the links given below.
RD Sharma Solutions for Class 11 Maths Exercise 16.3 Chapter 16 – Permutations
Also, access other exercises of RD Sharma Solutions for Class 11 Maths Chapter 16 – Permutations
Access answers to RD Sharma Solutions for Class 11 Maths Exercise 16.3 Chapter 16 – Permutations
1. Evaluate each of the following:
(i) 8P3
(ii) 10P4
(iii) 6P6
(iv) P (6, 4)
Solution:
(i) 8P3
We know that 8P3 can be 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 can be 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 can be 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
2. If P (5, r) = P (6, r – 1), find r.
Solution:
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 the obtained values in the 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)]
3. If 5 P(4, n) = 6 P(5, n – 1), find n.
Solution:
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 the obtained values in the 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)]
4. If P(n, 5) = 20 P(n, 3), find n.
Solution:
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 the obtained values in the 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)]
5. If nP4 = 360, find the value of n.
Solution:
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 the obtained values in the above expression, we get,
n!/(n – 4)! = 360
[n(n – 1) (n – 2) (n – 3) (n – 4)!] / (n – 4)! = 360n (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.
6. If P(9, r) = 3024, find r.
Solution:
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 the obtained values in the 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 r is 4.
7. If P (11, r) = P (12, r – 1), find r.
Solution:
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 the obtained values in the 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)]
8. If P(n, 4) = 12. P(n, 2), find n.
Solution:
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 the obtained values in the 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)]
9. If P(n – 1, 3) : P(n, 4) = 1 : 9, find n.
Solution:
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 the obtained values in the 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 n is 9.
10. If P(2n – 1, n) : P(2n + 1, n – 1) = 22 : 7 find n.
Solution:
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 the obtained values in the 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 = 22/77(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 n is 10.
11. If P(n, 5) : P(n, 3) = 2 : 1, find n.
Solution:
Given:
P(n, 5) : P(n, 3) = 2 : 1
P(n, 5) / P(n, 3) = 2 / 1
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) / P(n, 3) = 2 / 1
Substituting the obtained values in the above expression, we get,
[n!/ (n – 5)!] / [n!/ (n – 3)!] = 2/1 [n!/ (n – 5)!] × [(n – 3)! / n!] = 2/1(n – 3)! / (n – 5)! = 2/1
[(n – 3) (n – 3 – 1) (n – 3 – 2)!] / (n – 5)! = 2/1 [(n – 3) (n – 4) (n – 5)!] / (n – 5)! = 2/1(n – 3)(n – 4) = 2
n2 – 3n – 4n + 12 = 2
n2 – 7n + 12 – 2 = 0
n2 – 7n + 10 = 0
n2 – 5n – 2n + 10 = 0
n (n – 5) – 2(n – 5) = 0
(n – 5) (n – 2) = 0
n = 5 or 2
For, P (n, r): n ≥ r
∴ n = 5 [for, P (n, 5)]
12. Prove that:
1. P (1, 1) + 2. P (2, 2) + 3 . P (3, 3) + … + n . P(n, n) = P(n + 1, n + 1) – 1.
Solution:
By using the formula,
P (n, r) = n!/(n – r)!
P (n, n) = n!/(n – n)!
= n!/0!
= n! [Since, 0! = 1]
Consider LHS:
= 1. P(1, 1) + 2. P(2, 2) + 3. P(3, 3) + … + n . P(n, n)
= 1.1! + 2.2! + 3.3! +………+ n.n! [Since, P(n, n) = n!]
= (2! – 1!) + (3! – 2!) + (4! – 3!) + ……… + (n! – (n – 1)!) + ((n+1)! – n!)
= 2! – 1! + 3! – 2! + 4! – 3! + ……… + n! – (n – 1)! + (n+1)! – n!
= (n + 1)! – 1!
= (n + 1)! – 1 [Since, P (n, n) = n!]
= P(n+1, n+1) – 1
= RHS
Hence Proved.
13. If P(15, r – 1) : P(16, r – 2) = 3 : 4, find r.
Solution:
Given:
P(15, r – 1) : P(16, r – 2) = 3 : 4
P(15, r – 1) / P(16, r – 2) = 3/4
By using the formula,
P (n, r) = n!/(n – r)!
P (15, r – 1) = 15! / (15 – r + 1)!
= 15! / (16 – r)!
P (16, r – 2) = 16!/(16 – r + 2)!
= 16!/(18 – r)!
So, from the question,
P(15, r – 1) / P(16, r – 2) = 3/4
Substituting the obtained values in the above expression, we get,
[15! / (16 – r)!] / [16!/(18 – r)!] = 3/4 [15! / (16 – r)!] × [(18 – r)! / 16!] = 3/4 [15! / (16 – r)!] × [(18 – r) (18 – r – 1) (18 – r – 2)!]/(16×15!) = 3/41/(16 – r)! × [(18 – r) (17 – r) (16 – r)!]/16 = 3/4
(18 – r) (17 – r) = 3/4 × 16
(18 – r) (17 – r) = 12
306 – 18r – 17r + r2 = 12
306 – 12 – 35r + r2 = 0
r2 – 35r + 294 = 0
r2 – 21r – 14r + 294 = 0
r(r – 21) – 14(r – 21) = 0
(r – 14) (r – 21) = 0
r = 14 or 21
For, P(n, r): r ≤ n
∴ r = 14 [for, P(15, r – 1)]
14. n+5Pn+1 = 11(n – 1)/2 n+3Pn, find n.
Solution:
Given:
n+5Pn+1 = 11(n – 1)/2 n+3Pn
P (n +5, n + 1) = 11(n – 1)/2 P(n + 3, n)
By using the formula,
P (n, r) = n!/(n – r)!
(n + 5) (n + 4) = 22 (n – 1)
n2 + 4n + 5n + 20 = 22n – 22
n2 + 9n + 20 – 22n + 22 = 0
n2 – 13n + 42 = 0
n2 – 6n – 7n + 42 = 0
n(n – 6) – 7(n – 6) = 0
(n – 7) (n – 6) = 0
n = 7 or 6
∴ The value of n can either be 6 or 7.
15. In how many ways can five children stand in a queue?
Solution:
Number of arrangements of ‘n’ things taken all at a time = P (n, n)
So by using the formula,
By using the formula,
P (n, r) = n!/(n – r)!
The total number of ways in which five children can stand in a queue = the number of arrangements of 5 things taken all at a time = P (5, 5)
So,
P (5, 5) = 5!/(5 – 5)!
= 5!/0!
= 5! [Since, 0! = 1]
= 5 × 4 × 3 × 2 × 1
= 120
Hence, the number of ways in which five children can stand in a queue is 120.
16. From among the 36 teachers in a school, one principal and one vice-principal are to be appointed. In how many ways can this be done?
Solution:
Given:
The total number of teachers in a school = 36
We know the number of arrangements of n things taken r at a time = P(n, r)
By using the formula,
P (n, r) = n!/(n – r)!
∴ The total number of ways in which this can be done = the number of arrangements of 36 things taken 2 at a time = P(36, 2)
P (36, 2) = 36!/(36 – 2)!
= 36!/34!
= (36 × 35 × 34!)/34!
= 36 × 35
= 1260
Hence, the number of ways in which one principal and one vice-principal are to be appointed out of a total of 36 teachers in the school is 1260.
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