• +91 9971497814
• info@interviewmaterial.com

# RD Chapter 16- Permutations Ex-16.3 Interview Questions Answers

### Related Subjects

Question 1 :

Evaluate each of the following:

(i) 8P3

(ii) 10P4

(iii) 6P6

(iv) P (6, 4)

(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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

Todays Deals  ### RD Chapter 16- Permutations Ex-16.3 Contributors krishan

Name:
Email:

# Latest News # 9000 interview questions in different categories 