Permutations Notations :: Problems & Solutions

No. Problems & Solutions [Click Hide/Show to display the solutions below the question]
01.
Prove 0! = 1.

Solution » Hide/Show

nPr =
n!
(n − r)!

where r = n;
nPn =
n!
(n − n)!
n! =
n!
(0)!
0! =
n!
n!
0! = 1

02.
Find the value of 9P3.

Solution » Hide/Show

9P3 =
9!
(9 − 3)!
[
n!
(n − n)!
]
=
9 × 8 × 7 × 6!
6!
  = 9 × 8 × 7
  = 504

Alterative
9P3 = 9 × 8 × 7 [nPr = (n) (n − 1) ..... r times]
= 504

03.
Prove that P (10, 3 ) = P ( 9, 3 ) + 3 P ( 9, 2 )

Solution » Hide/Show

L.H.S =
=
=
P (10, 3)
10 × 9 × 8
720
R.H.S =
=
=
=
=
=
P (9, 3) + 3 P (9, 2)
9 × 8 × 7 + 3 × 9 × 8
9 × 8 (7 + 3)
10 × 9 × 8
720
L.H.S
04.
Find the value of 12P4 ÷ 10P3

Solution » Hide/Show

12P4 ÷ 10P3 =
12P4
10P3
=
12 × 11 × 10 .... 4 times
10 × 9 × .... 3 times
[nPr = (n) (n − 1) ...... to r times]
=
12 × 11 × 10 ×9
10 × 9 × 8
=
12 × 11
8
=
3 × 11
2
=
33
2
05.
Find the value of n if n P3 = 24.

Solution » Hide/Show

n P3 = 24

⇒ (n) (n − 1) ...... 3 times = 24 [nPr = (n) (n − 1) ..... r times]

⇒ (n) (n − 1) (n − 2) = 24 [(n) (n − 1) (n − 2) ⇒ Product of three consecutive numbers]

⇒ (n) (n − 1) (n − 2) = 4 × 3 × 2 [24 as a Product of three consecutive numbers ⇒ 24 = 4 × 3 × 2]

⇒ n = 4

06.
Find the value of 'n' if
np4
nP3
=
10
1
  (Or)   nP4 : nP3 = 10 : 1   (Or)   P (n, 4) = 10 P (n, 3)

Solution » Hide/Show

nP4 : nP3 = 10 : 1

nP4
nP3
=
10
1

nP4 = 10 nP3

(n) (n − 1) ... 4 times = 10 × (n) (n − 1) ... 3 times [nPr = (n) (n − 1) ... r times]

⇒ (n) (n − 1) (n − 2) (n − 3) = 10 × (n) (n − 1) (n − 2)

⇒ n − 3 = 10

⇒ n = 10 + 3

⇒ n = 13

07.
If nPn = 3,62,880, then n =

Solution » Hide/Show

nPn = 3,62,880

⇒ n! = 3,62,880 [ nPn = n!]

⇒ n! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9

⇒ n! = 9!

⇒ n = 9

08.
The number of permutations of 9 different things taking r at a time is 15,120. Then r =
(Or) Find "r" when 9Pr = 15,120

Solution » Hide/Show

The number of permutations of 9 different things taking r at a time is 15,120

9Pr = 15,120

⇒ 9 × 8 × 7 × ...... r times = 15,120

⇒ 9 × 8 × 7 × ...... r times = 9 × 8 × 7 × 6 × 5

⇒ 9 × 8 × 7 × ...... r times = 9 × 8 × 7 ... 5 times

⇒ r = 5

Alternative

9Pr = 15,120
9!
(9 − r)!
= 15,120 [
n!
(n − n)!
]
(9 − r)! =
9!
15,120
(9 − r)! =
3,62,880
15,120

⇒ (9 − r)! = 24

⇒ (9 − r) × (8 − r) × ..... × 1 = 24

⇒ (9 − r) × (8 − r) × ..... × 1 = 4 × 3 × 2 × 1
⇒ 9 − r = 4 (or) 8 − r = 3
⇒ r = 9 − 4   r = 8 − 3
⇒ r = 5   r = 5

09.
Find 'r' if 11P2r = 110   (Or)   P (11, 2r) = 110.

Solution » Hide/Show

11P2r = 110
11!
(11 − 2r)!
= 110 [
n!
(n − n)!
]
11 × 10 × 9!
(11 − 2r)!
= 110 [nPr = (n) (n − 1) ..... r times]

⇒ 110 × 9! = 110 × (11 − 2r)!

⇒ 9! = (11 − 2r)!

⇒ 11 − 2r = 9

⇒ 2r = 11 − 9

⇒ 2r = 2

⇒ r = 1

10.
If nPr = 120, then r =

Solution » Hide/Show

nPr = 120; If r = 1, nP1 = 120

⇒ n = 120

  If r = 2, nP2 = 120

⇒ (n) (n − 1) = 120

⇒ (n) (n − 1) = _____

[Since 120 cannot be written as a product of two consecutive integers,
r = 2 is inconsisten]

  If r = 3, nP3 = 120

⇒ (n) (n − 1) (n − 2) = 120

⇒ (n) (n − 1) (n − 2) = 6 × 5 × 4

⇒ n = 6

  If r = 4, nP4 = 120

⇒ (n) (n − 1) (n − 2) (n − 3) = 120

⇒ (n) (n − 1) (n − 2) (n − 3) = 5 × 4 × 3 × 2

⇒ n = 5

  If r = 5, nP5 = 120

⇒ (n) (n − 1) (n − 2) (n − 3) (n − 4) = 120

⇒ (n) (n − 1) (n − 2) (n − 3) (n − 4) = 5 × 4 × 3 × 2 × 1

⇒ n = 5

r = [1, 3, 4, 5],    (r, n) = [(1, 120) (3, 6) (4, 5) (5, 5)]

11.
Find the value of 'n' if
nP13
n + 1P12
=
3
4
  (Or)   P (n, 13) : P (n + 1, 12) = 3 : 4   (Or)   4 nP13 : 3 n + 1P12

Solution » Hide/Show

nP13 : n + 1P12 = 3 : 4
(n) (n − 1) ... × (n − 12)
(n + 1) (n) ... × (n − 10)
=
3
4
[(n) (n − 1) ... × (n − 10)] (n − 11) (n − 12)
(n + 1) [(n) ... × (n − 10)]
=
3
4
(n − 11) (n − 12)
(n + 1)
=
3
4
⇒ 4 (n2 − 12 n − 11 n + 132) = 3 (n + 1)
⇒ 4 (n2 − 23 n + 132) = 3 n + 3
⇒ 4 n2 − 92 n + 528 − 3 n − 3 = 0
⇒ 4 n2 − 95 n + 525 = 0
⇒ 4 n2 − 60 n − 35 n + 525 = 0
⇒ 4n (n − 15) − 35 (n − 15) = 0
⇒ (4n − 35) (n − 15) = 0
⇒ 4n − 35 = 0 or n − 15 = 0
⇒ 4n = 35 or n = 15
⇒ n =
35
4
or n = 15
Improbable

12.
Find the value of 'n' if four times the number of permutations of n things taken 3 together is equal to 2 times the number of permutations of (n - 1) things taken 3 together.

Solution » Hide/Show

Four times the number of permutations of n things taken 3 together is equal to 2 times the number of permutations of (n - 1) things taken 3 together.

nP3 = 2 × ( n − 1 )P3 [nPr = (n) (n − 1) ... r times]

⇒ (n) (n − 1) (n − 2) = 2 × (n − 1 ) ([n − 1] − 1) ([n − 1] − 2)

⇒ (n) (n − 1) (n − 2) = 2 × (n − 1 ) (n − 1 − 1) (n − 1 − 2)

⇒ (n) (n − 1) (n − 2) = 2 × (n − 1 ) (n − 2) (n − 3)

⇒ (n) = 2 × (n − 3)

⇒ n = 2n − 6

⇒ 6 = 2n − n

⇒ n = 6

13.
Find 'n' if np3 : (n + 2)P3 = 5 : 14

Solution » Hide/Show

np3 : (n + 2)P3 = 5 : 14

14 np3 = 5 (n + 2)P3

14 (n) (n − 1) (n − 2) = 5 (n + 2) [(n + 2) − 1] [(n + 2) − 2]

14 (n) (n − 1) (n − 2) = 5 (n + 2) (n + 2 − 1) (n + 2 − 2)

14 (n) (n − 1) (n − 2) = 5 (n + 2) (n + 1) (n)

14 (n − 1) (n − 2) = 5 (n + 2) (n + 1)

14 [n2 − 2n − n + 2] = 5 [n2 + n + 2n + 2]

14 [n2 − 3n + 2] = 5 [n2 + 3n + 2]

14 n2 − 42n + 28 = 5 n2 + 15n + 10

14 n2 − 5 n2 − 42n − 15n + 28 − 10 = 0

9 n2 − 57n + 18 = 0

9 n2 − 54n − 3n + 18 = 0

9n (n − 6) − 3 (n − 6) = 0

⇒ 9n − 3 = 0     (Or)     n − 6 = 0

⇒ n = 1/3 [inconsistent]     (Or)     n = 6

∴ n = 6

14.
Find n if (n + 3)p6 : (n + 2)P4 = 14 : 1

Solution » Hide/Show

(n + 3)p6 : (n + 2)P4 = 14 : 1

1 (n + 3)p6 = 14 (n + 2)P4

(n + 3) [(n + 3) − 1] [(n + 3) − 2] [(n + 3) − 3] [(n + 3) − 4] [(n + 3) − 5] = 14 (n + 2) [(n + 2) − 1] [(n + 2) − 2] [(n + 2) − 3]

(n + 3) (n + 3 − 1) (n + 3 − 2) (n + 3 − 3) (n + 3 − 4) (n + 3 − 5) = 14 (n + 2) (n + 2 − 1) (n + 2 − 2) (n + 2 − 3)

(n + 3) (n + 2) (n + 1) (n) (n − 1) (n − 2) = 14 (n + 2) (n + 1) (n) (n − 1)

(n + 3) (n − 2) = 14

Method I

(n + 3) (n − 2) = 14

⇒ (n + 3) (n − 2) = 7 × 2 [14 as a product of two numbers differing by 5]

⇒ n + 3 = 7     (Or)   n − 2 = 2

⇒ n = 7 − 3     (Or)   n = 2 + 2

⇒ n = 4     (Or)   n = 4

Method I

(n + 3) (n − 2) = 14

⇒ n2 + 3n − 2n − 6 = 14

⇒ n2 + n − 6 − 14 = 0

⇒ n2 + n − 20 = 0

⇒ n2 + 5n − 4n − 20 = 0

⇒ n (n + 5) − 4 (n + 5) = 0

⇒ (n − 4) (n + 5) = 0

⇒ n − 4 = 0     (Or)   n + 5 = 0

⇒ n = 4     (Or)   n = − 5 [inconsistent]

⇒ n = 4

15.
If 7 nP3 = 3 2nP2 then n =

Solution » Hide/Show

7 nP3 = 3 2nP2

⇒ 7 × (n) (n − 1) (n − 2) = 3 × (2n) (2n − 1)

⇒ 7 × (n − 1) (n − 2) = 3 × (2) (2n − 1)

⇒ 7 × (n2 − 2n − n + 2) = 6 × (2n − 1)

⇒ 7 × (n2 − 3n + 2) = 12n − 6

⇒ 7 n2 − 21n + 14 − 12n + 6 = 0

⇒ 7 n2 − 33n + 20 = 0

⇒ 7 n2 − 28n − 5n + 20 = 0

⇒ 7n (n − 4) − 5 (n − 4) = 0

⇒ (7n − 5) (n − 4) = 0

⇒ (7n − 5) = 0     (Or)     (n − 4) = 0

⇒ 7n = 5     (Or)     n = 4

⇒ n = 5/7 [inconsistent]     (Or)     n = 4

∴ n = 4.

16.
If ( n − 1 )P3 : ( n + 1 )P3 = 6 : 11, then n =

Solution » Hide/Show

( n − 1 )P3 : ( n + 1 )P3 = 6 : 11
( n − 1 )P3
( n + 1 )P3
=
6
11
( n − 1 ) ( n − 2) ( n − 3)
( n + 1 ) (n) ( n − 1 )
=
6
11
[nPr = (n) (n − 1) ............. r times]
( n − 2) ( n − 3)
( n + 1 ) (n)
=
6
11

⇒ 11 (n2 − 3 n − 2 n + 6) = 6 (n2 + n)

⇒ 11 n2 − 55 n + 66 = 6 n2 + 6n

⇒ 11 n2 − 6 n2 − 55 n − 6n + 66 = 0

⇒ 5 n2 − 61 n + 66 = 0

⇒ 5 n (n − 11) − 6 (n − 11) = 0

⇒ (5n − 6) (n − 11) = 0
⇒ 5 n − 6 = 0 (or) n − 11 = 0
⇒ 5 n = 6 (or) n = 11
⇒ n =
6
5
 
Inapproriate
∴ n = 11

17.
If (2n+1)P(n − 1) : (2n − 1)Pn= 3:5 then n =

Solution » Hide/Show

(2n+1)P(n − 1) : (2n − 1)Pn= 3:5

5 (2n+1)P(n − 1) = 3 (2n − 1)Pn
⇒ 5 ×
(2n+1)!
[(2n+1) − (n − 1)]
= 3 ×
(2n − 1)!
[(2n − 1) − n]!
5 × (2n+1) × (2n) × (2n − 1)!
(2n + 1 − n+ 1)!
= 3 ×
3 × (2n − 1)!
(2n − 1 − n)!
10n × (2n+1)
(n + 2)!
=
3
(n − 1)!

⇒ 10n × (2n+1) (n − 1)! = 3 (n + 2)!

⇒ 10n × (2n+1) (n − 1)! = 3 (n + 2) (n + 1) (n) (n − 1)!

⇒ 10 × (2n+1) = 3 (n + 2) (n + 1)

⇒ 20n + 10 = 3 (n2 + 2n + n + 2)

⇒ 20n + 10 = 3 (n2 + 3n + 2)

⇒ 20n + 10 = 3 n2 + 9n + 6

⇒ 3 n2 + 9n + 6 − 20n − 10 = 0

⇒ 3 n2 − 11n − 4 = 0

⇒ 3 n2 − 12n + n − 4 = 0

⇒ 3 n (n − 4) + 1 (n − 4) = 0

⇒ (3n + 1) (n − 4) = 0

⇒ 3n + 1 = 0     (Or)     n − 4 = 0

⇒ 3n = − 1 [inconsistent]     (Or)     n = 4

∴ n = 4.

18.
If 6pr : 6 P5 = 1 : 2 find the value of r

Solution » Hide/Show

6pr : 6 P5 = 1 : 2

⇒ 2 6pr = 6 P5

⇒ 2 6pr = 6 × 5 × 4 × 3 × 2

6pr = 6 × 5 × 4 × 3

6pr = 6 × 5 × ... 4 times

⇒ 6 × 5 × ... r times = 6 × 5 × ... 4 times

⇒ r = 4

Alternative

6pr : 6 P5 = 1 : 2

2 6pr = 6 P5
⇒ 2 ×
(6)!
(6 − r)!
=
(6)!
(6 − 5)!
⇒ 2 ×
1
(6 − r)!
=
1
(1)!
2
(6 − r)!
= 1

⇒ (6 − r)! = 2

⇒ (6 − r)! = 2 × 1

⇒ (6 − r)! = 2!

⇒ 6 − r = 2

⇒ 6 − 2 = r

⇒ r = 4

19.
If 7P6 = 24 7P(r − 1) then r =

Solution » Hide/Show

7P6 = 24 7P(r − 1)
(7)!
(7 − 6)!
= 24 ×
(7)!
(7 − (r − 1))!
1
(1)!
=
24 × 1
(7 − r + 1)!

⇒ (8 − r)! = 24

⇒ (8 − r)! = 4 × 3 × 2 × 1

⇒ (8 − r)! = 4!

⇒ 8 − r = 4

⇒ 8 − 4 = r

⇒ r = 4

20.
Find 'r' if 4 6Pr = 6P( r + 1 )

Solution » Hide/Show

4 6Pr = 6P( r + 1 )
⇒ 4 ×
(6)!
(6 − r)!
=
(6)!
(6 − (r + 1) )!
4
(6 − r)!
=
1
(6 − r − 1)!

⇒ 4 (5 − r)! = (6 − r)!

⇒ 4 (5 − r)! = (6 − r) (5 − r)!

⇒ 4 = 6 − r

⇒ r = 6 − 4

⇒ r = 2

21.
Find 'r' if P (10, r + 1 ) : P ( 11, r ) = 30 : 11

Solution » Hide/Show

P (10, r + 1 ) : P ( 11, r ) = 30 : 11
P (10, r +1)
P (11, r)
=
30
11
10!
(10 − (r +1))!
11!
(11 − r)!
=
30
11
10!
(10 − r − 1)!
×
(11 − r)!
(11 )!
=
30
11
10!
(9 − r)!
×
(11 − r)!
11 × 10!
=
30
11
(11 − r) (10 − r) (9 − r)!
(9 − r)!
= 30
(11 − r) (10 − r) = 30
Method I
⇒ (11 − r) (10 − r) = 6 × 5
⇒ 11 − r = 6
⇒ r = 11 − 6
⇒ r = 5
Method II
(110 − 11 r &minus 10 r + r2) = 30
r2 − 21 r + 110 − 30 = 0
r2 − 21 r + 80 = 0
r2 − 16 r − 5 r + 80 = 0
r (r − 16) − 5 (r − 16) = 0
r − 5 = 0 (or) r − 16 = 0
r = 5 (or) r = 16
Since r cannotbe greater than n
r ≠ 16
r = 5

22.
Find 'r' if 5 4Pr = 6 5P(r − 1)     (Or) If P(4, r) : P(5, r − 1) = 6 : 5, then r = ?

Solution » Hide/Show

5 4Pr = 6 5P( r − 1 )
⇒ 5 ×
(4)!
(4 − r)!
= 6 ×
(5)!
(5 − (r − 1) )!
5 × 4!
(4 − r)!
=
6 × 5 × 4!
(5 − r + 1)!
1
(4 − r)!
=
6
(6 − r)!

⇒ (6 − r)! = 6 (4 − r)!

⇒ (6 − r) (5 − r) (4 − r)! = 6 (4 − r)!

⇒ (6 − r) (5 − r) = 6

Method I

⇒ (6 − r) (5 − r) = 6

⇒ (6 − r) (5 − r) = (3) (2)
⇒ 6 − r = 3 or 5 − r = 2
⇒ r = 6 − 3 or r = 5 − 2
⇒ r = 3 or r = 3

Method II

⇒ (6 − r) (5 − r) = 6

⇒ 30 − 6 r −5 r + r2 = 6

⇒ 30 − 11 r + r2

⇒ r2 − 11 r + 30 − 6 = 0

⇒ r2 − 11 r + 24 = 0

⇒ r2 − 8 r − 3 r + 24 = 0

⇒ r(r − 8) − 3 (r − 8) = 0

⇒ (r − 3) (r − 8) = 0
r − 3 = o (or) r − 8 = 0
r = 3 (or) r = 8
r ≠ 8 [Since r cannot be greater than n (4 on LHS or 5 on RHS)]
∴ r =3

23.
If 18p(r − 1) : 17 P(r − 1) = 9 : 7 then r.

Solution » Hide/Show

18p(r − 1) : 17 P(r − 1) = 9 : 7

⇒ 7 18p(r − 1) = 9 17 P(r − 1)
⇒ 7 ×
(18)!
[18 − (r − 1)]!
= 9 ×
(17)!
[17 − (r − 1)]!
7 × 18 × 17!
(18 − r + 1)!
=
9 × 17!
(17 − r + 1)!
7 × 2
(19 − r)!
=
1
(18 − r)!

⇒ 14 (18 − r)! = 1 (19 − r) × (18 − r)!

⇒ 14 = 19 − r

⇒ r = 19 − 14

⇒ r = 5

24.
If 10p(r − 1) : 11 P(r − 2) = 30 : 11 find r.

Solution » Hide/Show

P ( 11, r ) = P ( 12, r − 1 )
11!
(11 − r)!
=
12!
(12 − (r − 1))!
11!
(11 − r)!
=
12 × 11!
(12 − r + 1)!
1
(11 − r)!
=
12
(13 − r)!
⇒ (13 − r) (12 − r) (11 − r)! = 12 (11 − r)!
⇒ (13 − r) (12 − r) = 12
Method I
⇒ (13 − r) (12 − r) = 4 × 3
⇒ 13 − r = 4
⇒ r = 13 − 4
⇒ r = 9
Method II
156 − 13 r &minus 12 r + r2 = 12
⇒ 156 − 25 r + r2) = 12
⇒ r2 − 25 r + 156 − 12 = 0
⇒ r2 − 25 r + 144 = 0
⇒ r2 − 9 r − 16 r + 144 = 0
⇒ r (r − 9) − 16 (r − 9) = 0
⇒ (r − 9) (r − 16) = 0
⇒ r − 9 = 0 (or) r − 16 = 0
⇒ r = 9 (or) r = 16
Since r cannot be greater than 'n' (12 or 11)
r ≠ 16
∴ r = 9

25.
Find r if 7Pr = 60 7 P(r − 3)

Solution » Hide/Show

7Pr = 60 7 P(r − 3)
7!
(7 − r)!
= 60 ×
7!
[7 − (r − 3)]!
1
(7 − r)!
=
60 × 1
(7 − r + − 3)!

⇒ (10 − r)! = 60 (7 − r)!

⇒ (10 − r) × (9 − r) × (8 − r) × (7 − r)! = 60 (7 − r)!

⇒ (10 − r) × (9 − r) × (8 − r) = 60

⇒ (10 − r) × (9 − r) × (8 − r) = 5 × 4 × 3

⇒ 10 − r = 5   (Or)   9 − r = 4   (Or)   8 − r = 3

⇒ 10 − 5 = r   (Or)   9 − 4 = r   (Or)   8 − 3 = r

⇒ r = 5

26.
If 56 P ( r + 6 ) : 54 P ( r + 3 ) = 30,800 : 1, find r   (Or)     P(56, r + 6) = 30,800 P(54, r + 3). Find r.

Solution » Hide/Show

56 P ( r + 6 ) : 54 P ( r + 3 ) = 30800 : 1
56 P ( r + 6 )
54 P ( r + 3 )
=
30800
1
56!
(56 − (r + 6))!
54!
(54 − (r + 2))!
=
30800
1
56!
(56 − r − 6)!
×
(54 − r − 3)!
(54)!
= 30800
56 × 55 × (54)! × (51 − r)!
(50 − r)! × (54)!
= 30800
56 × 55 × (51 − r) (50 − r)!
(50 − r)!
= 30800
⇒ 56 × 55 × (51 − r) = 30800
(51 − r) =
30800
56 × 55
⇒ (51 − r) = 10
⇒ r = 51 − 10
⇒ r = 41

27.
If m+nP2 = 56 and m-nP2=12 find m and n.

Solution » Hide/Show

m + nP2 = 56
aP2 = 56
a!
(a − 2)!
= 56
(a) (a − 1) (a − 2)!
(a − 2)!
= 56
(a) (a − 1) = 56
(a) (a − 1) = 8 × 7
a = 8       → (3)

m − nP2 = 12
bP2 = 12
(b)!
(b − 2)!
= 12
(b)(b − 1) (b − 2)!
(b − 2)!
= 12
(b) (b − 1) = 12
(b) (b − 1) = 4 × 3
b = 4           →   (4)

(3) → a = 8 m + n = 8       →  (5)
(4) → b = 4 m − n = 4       →  (6)
(5) + (6)    2 m = 12
  m =
12
2
  m = 6
(5) ⇒ m + n = 8
  ⇒ 6 + n = 8
  ⇒ n = 8 − 6
  ⇒ n = 2
∴ (m , n) = (6 , 2)

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No. Problems for Practice
01.
A.  
02.
A. Evaluate 10 P4 B. Evaluate 15 P0 C. What is the value of 10P3
D. Find the value of78 P3 E. Determine the value of 40P3 F. Find 7P7
G. Evaluate 9 P1 H. Evaluate 9 P3 + 5
03.
A.  
04.
A. Evaluate 15P4 ÷ 12P3
05.
A. Find the value of 'n' if nP4 = 360. B. What is the value of 'n' when nP2 = 12.
C. nP3 = 1320, then n = D. Find 'n' if nP3 = 720.
06.
A. Find the value of 'n' if nP5 = 42 nP3 (n≥4) B. For what value of 'n' is nP5 = 20 nP3
C. nP5 = nP6, then n = D. If nP7 = 42 nP5, then n =
E. If nP6 = 30 nP4 find the value of n F. Solve for n given nP4 = 30 nP2
G. If np5 : nP3 = 2 : 1 Find 'n'. H. If np4 : nP6 = 9 : 1 Find the value of 'n'.
I.
Solve for n given
np5
nP3
=
2
1
J.
07.
A. Find n if nPn = 40,320 B. nPn = 720, then n=
08.
A. If 10Pr = 5,040, then r = B. Find 'r' if 8Pr = 1,680.
C. The number of permutations of 10 different things taking r at a time is 1,51,200. Then r =
09.
A.  
10.
A. Find n and r if nPr = 7,920 B. If nPr = 1,716, then (n, r) =
C. If nPr = 3,024, then r = D. nPr = 1,680, r =
11.
A. Find the value of 'n' if 16 nP3 = 13 n + 1P3
A. Find n if ( n + 1 )p4 : n P5 = 2 : 3
D. If (n + 1)p5 : nP6 = 2 : 7 then n = .
12.
A. If 3 np3 = 4 ( n -1 )P3, then n =
B. If 6 np3 = ( n -1 )P4, then n =
C. Find n if 20 np4 = 9 ( n - 1 )P5
D. Find n if np4 = 2 ( n - 1 )P4
B. If n P 5 : (n - 1) P 5 = 3 : 2 then n =
13.
A. Find the value of 'n' such that 30 nP6 = ( n + 2)P7 (n ≥ 6)
14.
15.
16.
A. Find n if (n − 1)p3 : (n + 1)P3 = 5 : 12
17.
18.
19.
A. Find r if 7p6 = 6 7 P( r - 1 )
20.
21.
22.
A. If P (11, r ) = P (12, r − 1 ) find r
B. If 20pr = 13 20 P( r - 1 ) then r =
C. If 20p r : 20 P( r - 1 ) = 15 : 1 find r =
D. 12pr : 12 P( r - 1 ) = 6 : 1 find r.
E. If 20 P r : 20 P (r - 1) = 15 : 1 then what is 'r'
23.
A. If 15p( r - 1 ) : 16 P( r - 2 ) = 3 : 4 then r = .
24.
25.
26.
27.

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