Combinations Notations :: Problems & Solutions

No. Problems & Solutions [Click Hide/Show to display the solutions below the question]
01.
Define combination? What is the relationship between nCr and nPr.

Solution » Hide/Show

 

02.
nCr =
nPr
r!

Solution » Hide/Show

L.H.S =
n!
(n − r)! r!
R.H.S =
nPr
r!
 
=
n!
(n − r)!
r!
  =
n!
(n − r)!
×
1
r!
  =
n!
(n − r)! r!
  = L.H.S

03.
Find the value of 6C3

Solution » Hide/Show

6C3 =
6!
(6 − 3)! 3!
[ nCr =
n!
(n − r)! r!
]
  =
6 × 5 × 4 × 3!
3! × 3!
  =
6 × 5 × 4
3× 2 × 1
  =
=
5 × 4
20

Alt
6C3 =
6 × 5 × 4
3!
[ nCr =
(n) (n − 1) ....... r times
r!
]
  =
6 × 5 × 4
3× 2 × 1
  =
=
5 × 4
20

04.
Find n if nC2 = 36.

Solution » Hide/Show

nC2 = 36.
n (n − 1)
2!
= 36 [ nCr =
n!
(n − r)! × r!
]

⇒ n (n − 1) = 36 × 2

⇒ n (n − 1) = 72

Method I

⇒ n (n − 1) = 72

⇒ n (n − 1) = 9 × 8

⇒ n = 9   (Or) (n − 1) = 8

⇒ n = 9   (Or) n = 8 + 1

⇒ n = 9

Method II

⇒ n (n − 1) = 72

⇒ n2 − n = 72

⇒ n2 − n − 72 = 0

⇒ n2 − 9n + 8n − 72 = 0

⇒ n (n − 9) + 8 (n − 9) = 0

⇒ (n − 8) (n − 9) = 0

⇒ n + 8 = 0   (Or)   n − 9 = 0

⇒ n = − 8 [inconsistent]   (Or)   n = 9

Therefore, n = 9.

05.
nCr = nC n − r .

Solution » Hide/Show

L.H.S = nCr
  =
n!
(n − r)! r!
R.H.S = nCn − r
  =
n!
(n −(n − r) )! (n − r)!
  =
n!
(n − n + r) )! (n − r)!
  =
n!
r! (n − r)!
  = L.H.S

06.
Find the value of nCn − 3

Solution » Hide/Show

nCn − 3 =
n!
(n −(n − 3) )! (n − 3)!
[ nCr =
n!
(n − r)! × r!
]
  =
n!
(n − n + 3)! (n − 3)!
  =
n!
3! (n − 3)!
  =
n (n − 1) (n − 2) (n − 3)!
3 × 2 × 1 × (n − 3)!
  =
1
6
(n) (n − 1) (n − 2)

Alternative
nC(n − 3) = nC[n − (n − 3)] [nCr = nC(n − r)]
= nCn − n + 3
= nC3
=
(n) × (n × 1) × (n × 2)
3!
=
(n) × (n × 1) × (n × 2)
3 × 2 × 1
  =
1
6
(n) (n − 1) (n − 2)

07.
Find n if nCn − 4 = 15.

Solution » Hide/Show

nCn − 4 = 15.
n!
(n − (n − 4))! (n − 4)!
= 15 [ nCr =
n!
(n − r)! × r!
]
n!
(n − n + 4)! (n − 4)!
= 15
n (n − 1) (n − 2) (n − 3) (n − 4)
4! × (n − 4)!
= 15

⇒ n (n − 1) (n − 2) (n − 3) = 15 × 24

⇒ n (n − 1) (n − 2) (n − 3) = 360

⇒ n (n − 1) (n − 2) (n − 3) = 6 × 5 × 4 × 3

⇒ n = 6   (Or) n − 1 = 5   (Or) n − 2 = 4   (Or) n − 3 = 3

⇒ n = 6

Alternative

nCn − 4 = 15.

nCn − (n − 4) = 15 [nCr = nCn − r]

nCn − n + 4 = 15

nC4 = 15

(n) (n − 1) (n − 2) (n − 3)
4!
= 15 [ nCr =
n × (n − 1) × (n − 2) ... r times
r!
]

⇒ n (n − 1) (n − 2) (n − 3) = 15 × 24

⇒ n (n − 1) (n − 2) (n − 3) = 360

⇒ n (n − 1) (n − 2) (n − 3) = 6 × 5 × 4 × 3

⇒ n = 6   (Or) n − 1 = 5   (Or) n − 2 = 4   (Or) n − 3 = 3

⇒ n = 6

08.
If n + 1C4 = 9 nC2, find the value of 'n'.

Solution » Hide/Show

n + 1C4 = 9 nC2

(n + 1) × (n) × (n − 1) × (n − 2)
4!
= 9 ×
n × (n − 1)
2!
(n + 1) × (n − 2)
4 × 3 × 2 × 1
= 9 ×
1
2 × 1

⇒ (n + 1) × (n − 2) = 9 × 12

Method I [From this step]

⇒ (n + 1) × (n − 2) = 12 × 9

⇒ n + 1 = 12   (Or)   n − 2 = 9

⇒ n = 12 − 1   (Or)   n = 9 + 2

⇒ n = 11

Method II

⇒ n2 − 2n + n − 2 = 108

⇒ n2 − n − 2 − 108 = 0

⇒ n2 − n − 110 = 0

⇒ n2 − 11n + 10n − 110 = 0

⇒ n (n − 11) + 10 (n − 11) = 0

⇒ (n + 10) (n − 11) = 0

⇒ n + 10 = 0   (Or)   n − 11 = 0

⇒ n = − 10 [inconsistent]   (Or)   n = 11

⇒ n = 11

09.
If nCr = 126 find n.

Solution » Hide/Show

nCr = 126
⇒ r = 0 → nC0 = 126 not relevant since nC0 = 1
  r = 1 → nC1 = 126 ⇒
(n)!
1!
= 126 ⇒ n = 126
  r = 2 &rarr nC2 = 126 ⇒
(n) (n − 1)
2!
= 126
  ⇒ (n) (n − 1) = 126
  ⇒ 17 × 16 = 272; 16 × 15 = 240 ; → Not Possible
  r = 3 &rarr nC3 = 126 ⇒
(n) (n − 1) (n − 2)
3!
= 126
  ⇒ (n) (n − 1) (n − 2) = 126 × 3!
  = 126 × 6
  r = 4 &rarr nC4 = 126 ⇒
(n) (n − 1) (n − 2) (n − 3)
4!
= 126
  ⇒ (n) (n − 1) (n − 2) (n − 3) = 126 × 24
⇒ (n) (n − 1) (n − 2) (n − 3) = 9 × 8 × 7 × 6
⇒ n = 9
∴ (n, r) = (126, 1) or (9, 4)

10.
If nC6 = nC10 find n and evaluate nC4

Solution » Hide/Show

nC6 = nC10
n!
(n − 6)! 6!
=
n!
(n − 10)! 10!
[ nCr =
n!
(n − r)! × r!
]

⇒ 6! (n − 10)! = (n − 6) (n − 7) (n − 8) (n − 9) (n − 10)! 6!

⇒ 10 × 9 × 8 × 7 × 6! = (n − 6) (n − 7) (n − 8) (n − 9) × 6!

⇒ (n − 6) (n − 7) (n − 8) (n − 9) = 10 × 9 × 8 × 7

⇒ (n − 6) = 10   (Or) (n − 7) = 9   (Or) ...

⇒ n = 10 + 6

⇒ n = 16

Alternative I

nC6 = nC10

We know nC6 = nCn − 6 [nCr = nCn − r]

From the above two we can say nCn − 6 = nC10

⇒ n − 6 = 10

⇒ n = 10 + 6

⇒ n = 16

Alternative II

nC6 = nC10

⇒ n = 6 + 10   (Or)   6 = 10 [Inconsistent] [nCa = nCb ⇒ a = b (Or) a + b = n]

⇒ n = 16

Value of nC4 = 16C4
=
16 × 15 × 14 × 13
4!
=
16 × 15 × 14 × 13
4 × 3 × 2 × 1
= 2 × 5 × 14 × 13
= 1,820

11.
Find r if 20C2r − 3 = 20C2r − 1

Solution » Hide/Show

20C2r − 3 = 20C2r − 1

⇒ 20 = (2r − 3) + (2r − 1)   (Or)   2r − 3 = 2r − 1 [Inconsistent] [nCa = nCb ⇒ n = a + b   (Or) a = b]

20 = 2r − 3 + 2r − 1

⇒ 20 = 4r − 4

⇒ 20 + 4 = 4r

⇒ 4r = 24

⇒ r = 6

12.
If 2nCr = 2nCr + 2 find the value of r

Solution » Hide/Show

2nCr = 2nCr + 2

⇒ 2n = r + (r + 2)   (Or)   r = r + 2 [inconsistent] [nCa = nCb ⇒ n = a + b   (Or) a = b]

⇒ 2n = 2r + 2

⇒ 2n = 2(r + 1)

⇒ n = r + 1

⇒ r = n − 1

13.
Find r if 42Cr + 3 = 42C2r − 5

Solution » Hide/Show

42Cr + 3 = 42C2r − 5

⇒ 42 = (r + 3) + (2r − 5)   (Or)   r + 3 = 2r − 5 [nCa = nCb ⇒ n = a + b   (Or) a = b]

⇒ 42 = r + 3 + 2r − 5   (Or)   3 + 5 = 2r − r

⇒ 42 = 3r − 2   (Or)   8 = r

⇒ 42 + 2 = 3r   (Or)   r = 8

⇒ 3r = 44   (Or) ...

⇒ r = 44/3 [inconsistent]   (Or) ...

⇒ r = 8

14.
Find n if nC3 : n − 1C4 = 8 : 5

Solution » Hide/Show

nC3 : n − 1C4 = 8 : 5
nC3
n − 1C4
=
8
5
(n) (n − 1) (n − 2)
3!
(n − 1) (n − 2) (n − 3) (n − 4)
4!
=
8
5
(n) (n − 1) (n − 2)
3!
×
4!
(n − 1) (n − 2) (n − 3) (n − 4)
=
8
5
(n) × 4 × 3!
(n − 3) (n − 4) × 3!
=
8
5
n
(n − 3) (n − 4)
=
2
5

⇒ 5n = 2 (n − 3) (n − 4)

⇒ 5n = 2 (n2 − 3 n − 4 n + 12)

⇒ 5n = 2 (n2 − 7 n + 12)

⇒ 5n = 2 n2 − 14 n + 24

⇒ 2 n2 − 14 n − 5n + 24= 0

⇒ 2 n2 − 19 n + 24= 0

⇒ 2 n2 − 16 n − 3n + 24= 0

⇒ 2 n (n − 8) − 3 (n − 8 ) = 0

⇒ (2 n − 3) (n − 8) = 0

⇒ 2n − 3 = 0   (Or)   n − 8 = 0

⇒ n = 3/2 [Inconsistent]   (Or)   n = 8

⇒ n = 8

15.
If 2nC3 : 2nC2 = 8 find n

Solution » Hide/Show

2nC3 : 2nC2 = 8
2nC3
2nC2
= 8
(2n) (2n − 1) (2n − 2)
3!
(2n) (2n − 1)
2!
= 8
(2n) (2n − 1) (2n − 2)
3 × 2 × 1
×
2 × 1
(2n) (2n − 1)
= 8
2n − 2
3
= 8

⇒ 2 (n − 1) = 3 × 8

⇒ n − 1 = 12

⇒ n = 12 + 1

⇒ n = 13

16.
2nC2 − 2 nC2 =

Solution » Hide/Show

2nC2 − 2 nC2 =
(2n) (2n − 1)
2!
− 2 ×
(n) (n − 1)
2!
=
[(2n) (2n − 1)] − 2 [(n) (n − 1)]
2!
=
2n [(2n − 1) − (n − 1)]
2 × 1!
=
2n (2n − 1 − n + 1)
2
= n × n
= n2

17.
If P = mC2 , find pC2

Solution » Hide/Show

P = mC2
⇒ P =
(m) (m − 1)
2 × 1
→ (1)
PC2 =
(p) (p − 1)
2 × 1
 

=
(m) (m − 1)
2 × 1
[ (m) (m − 1)
2 × 1
− 1]
1
2 × 1
  =
(m) (m − 1)
2
×
(m) (m − 1) − 2
2
×
1
2
  =
1
8
(m) (m − 1) [m2 − m − 2]
  =
1
8
(m) (m − 1) [m2 − 2m + m − 2]
  =
1
8
(m) (m − 1) [m (m − 2) + 1 (m − 2) ]
  =
1
8
(m) (m − 1) [ (m + 1) (m − 2) ]
  =
1
8
(m + 1) (m) (m − 1) (m − 2]

18.
If 3nCr = 3n Cn + r express the values of each in term of n.

Solution » Hide/Show

3nCr = 3n Cn + r

⇒ 3n = r + (n + r)   (Or)   r = n + r [inconsistent] [nCa = nCb ⇒ n = a + b   (Or) a = b]

⇒ 3n = 2r + n

⇒ 3n − n = 2r

⇒ 2n = 2r

⇒ n = r

3nCr =
3n!
(3n − r)! × r!
=
3n!
(3n − n)! × n!
=
3n!
(2n)! × n!

3n Cn + r =
3n!
(3n − [n + r])! × (n + r)!
=
3n!
(3n − [n + n])! × (n + n)!
=
3n!
(3n − 2n)! × (2n)!
=
3n!
n! × (2n)!

19.
If n and r are non- negative integers, such that 1 ≤ r ≤ n, then prove that nCr + n Cr − 1 = n + 1 Cr

Solution » Hide/Show

L.H.S = nCr + nCr − 1
  =
n!
(n − r)! r!
+
n!
[n − ( r − 1)]! (r − 1)!
  =
n!
(n − r)! r!
+
n!
(n − r + 1)! (r − 1)!
[L.C.M of r!, (r − 1)! is r!]
[L.C.M of (n − r + 1)!, (n − r)! is (n − r + 1)!]
  =
(n − r + 1) n! + (r) (n!)
(n − r + 1)! r!
(n − r + 1)!
(n − r)! r!
= (n − r + 1)
  =
n! [n − r + 1 + r ]
(n − r + 1)! r!
r!
(r − 1)!
= r
  =
n! (n + 1)
[(n + 1) − r]! r!
  =
(n + 1)!
[(n + 1) − r]! r!
  = n + 1Cr
  = R.H.S

This is better remembered as n Cr − 1 + nCr = n + 1 Cr

20.
Prove that 20C13 + 20C14 = 20C6 + 20C7

Solution » Hide/Show

L.H.S = 20C13 + 20C14
  =
20!
(20 − 13)! 13!
+
20!
(20 − 14)! 14!
  =
20!
7! 13!
+
20!
6! 14!
  =
14 × 20! + 7 × 20!
14! 7!
  =
(14 + 7) 20!
14! 7!
  =
21 × 20!
14! 7!
  =
21!
14! 7!
  = 21C7
R.H.S = 20C6 + 20C7
  =
20!
(20 − 6)! 6!
+
20!
(20 − 7)! 7!
  =
20!
14! 6!
+
20!
13! 7!
  =
7 × 20! + 14 × 20!
14! 7!
  =
(7 + 14) 20!
14! 7!
  =
21 × 20!
14! 7!
  =
21!
14! 7!
  = 21C7

∴ L.H.S = R.H.S

Alternative I
20C13 + 20C14 =20C(20 − 13) + 20C(20 − 14)
  =20C7 + 20C6
  = R.H.S

Alternative II
LHS = 20C13 + 20C14
= 21C14 [nC(r − 1) + nCr = (n + 1)Cr
RHS = 20C7 + 20C6
= 21C7 [ nCr + nC(r − 1) = (n + 1)Cr]
= 21C(21 − 7) [nCr = nC(n − r)]
= 21C14

⇒ L.H.S = R.H.S

21.
Evaluate 15C8 + 15C915C615C7

Solution » Hide/Show

15C8 + 15C915C615C7 = 15C(15 − 8) + 15C(15 − 9) − (15C6 + 15C7)
  = 15C7 + 15C6 − (15C6 + 15C7)
  = (15C7 + 15C6) − (15C6 + 15C7)
  = 0

Alternative
15C8 + 15C915C615C7 = (15C9 + 15C8) − (15C7 + 15C6)
= 16C916C7 [ nCr + nC(r − 1) = (n + 1)Cr]
= 16C916C(16 − 7) [nCr = nC(n − r)]
= 16C916C9
= 0

22.
If 8Cr7C3 = 7C2 Find r

Solution » Hide/Show

8Cr7C3 = 7C2

8Cr = 7C2 + 7C3
8Cr =
7 × 6
2 × 1
+
7 × 6 × 5
3 × 2 × 1

8Cr = 21 + 35

8Cr = 56

8Cr = 8 × 7
8Cr =
8 × 7 ×
3!
3!
8Cr =
8 × 7 × 6
3!
8Cr = 8C3

⇒ r = 3

Method II

8Cr7C3 = 7C2

8Cr = 7C2 + 7C3

8Cr = 8C3 [nC(r − 1) + nCr = (n + 1)Cr]

⇒ r = 3

23.
Find n if n + 1C3 = 2 nC2

Solution » Hide/Show

n + 1C3 = 2 nC2
(n +1) (n) (n − 1)
3!
= 2 ×
(n) (n − 1)
2!
(n +1)
6
=
2
2 × 1
n + 1 = 6
n = 6 − 1
n = 5

Alternative

n + 1C3 = 2 nC2

nC3 + nC2= 2 nC2 [(n + 1)Cr = nCr + nC(r − 1)]

nC3 = 2 nC2nC2

nC3 = nC2

⇒ n = 3 + 2 [nCa = nCb ⇒ n = a + b   (Or) a = b]

⇒ n = 5

24.
If 8C3 + (n + 2)C4 = 9C4 then n =

Solution » Hide/Show

8C3 + (n + 2)C4 = 9C4

(n + 2)C4 = 9C48C3

(n + 2)C4 = 8C4 + 8C38C3 [(n + 1)Cr = nCr + nC(r − 1)]

(n + 2)C4 = 8C4

⇒ n + 2 = 8

⇒ n = 8 − 2

⇒ n = 6

25.
Prove that nCr + 2 nCr − 1 + nCr − 2 = n + 2Cr

Solution » Hide/Show

L.H.S = nCr + 2 nCr − 1 + nCr − 2
  = nCr + nCr − 1 + nCr − 1+ nCr − 2
  = (nCr + nCr − 1 ) + (nCr − 1+ nCr − 2)
  = (n + 1Cr) + (n + 1Cr − 1) [nCr + nCr − 1 = n + 1Cr)
  = ([n + 1] + 1)Cr [nCr + nCr − 1 = n + 1Cr)
  = (n + 1 + 1)Cr
  = n + 2Cr
  = R.H.S

26.
Find the value of 8C4 + 2 8C3 + 8C2

Solution » Hide/Show

8C4 + 2 8C3 + 8C2 = (8C4 + 8C3) + (8C3 + 8C2)
= 9C4 + 9C3 [nCr + nCr − 1 = n + 1Cr)
= 10C4 [nCr + nCr − 1 = n + 1Cr)
=
10 × 9 × 8 × 7
4!
=
10 × 9 × 8 × 7
4 × 3 × 2 × 1
= 10 × 3 × 7
= 210

27.
Find x if 12C5 + 2 12C4 + 12C3 = 14Cx

Solution » Hide/Show

12C5 + 2 12C4 + 12C3 = 14Cx

12C5 + 12C4 + 12C4 + 12C3 = 14Cx

13C5 + 13C4 = 14Cx

14C5 = 14Cx

⇒ x = 5

28.
2nCn + 1 + 2 2nCn + 2nCn − 1 =

Solution » Hide/Show

2nCn + 1 + 2 2nCn + 2nCn − 1 = 2nCn + 1 + 2nCn + 2nCn + 2nCn − 1
= (2n + 1)Cn + 1 + (2n + 1)Cn
= ((2n + 1) + 1)Cn + 1
= (2n + 1 + 1)Cn
= (2n + 2)Cn
= 2(n + 1)Cn

29.
If nCr + n − 1 Cr − 1 + n − 1 Cr − 2 = n + 1 Cr

Solution » Hide/Show

LHS = nCr + n − 1 Cr − 1 + n − 1 Cr − 2
= nCr + nCr − 1
= n + 1Cr
= RHS

permutations,combinations,linear,circular,counting,principles,probability, permutations,combinations,linear,circular,counting,principles,probability,
No. Problems for Practice
01.
02.
03.
A. Find the value of 15C4 B. How much is 21C19 = C. 8C5 = ?
D. Find the value of 9C0 E. Evaluate 100C2 F. What is the value of 7C1
G. Find the value of 35C32 H. Evaluate 14C14 I. Find the value of 10C8
04.
A. If nC3 = 84 then n = B. Find n if nC2 = 45 C. Find n ifnC4 = 126
05.
 
06.
Prove that nC2 =
1
2
(n) (n − 1)
07.
A. Find n in the following nCn − 3 = 10
08.
E. n + 2Cn = 45 find n F. 4 nC2 = n + 2C3 find n
09.
 
10.
A. Find n if nC20 = nC35 B. nC10 = nC15 find the value of 27Cn
C. nC16 = nC9 find nC23 and 27Cn D. 11Cx = 11Cyand x ≠y what is the value of x + y
E. If C(n,8) = C(n,6) find C(n,2) F. n C10 = nC8, 21Cn =
11.
A. If 18Cr = 18Cr + 2, find the value of 'r' and rC3
B. If C(10,r) = C(10, r + 4) Find the value of C(5, r)
C. 15C2r + 3 = 15C3r + 2 find 10Cr
D. 36C2r − 3 = 36C2r − 1 then r =
E. If 35Cn + 7 = 35C4n − 2 then the values of n are given by ____.
F. Find r in the following 15C(r + 3) = 15C(2r − 3)
G. Find r in the following 20C3n = 20C2n + 5
H. 20C2r − 3 = 20C3r − 2 then r =
12.
A. If nCr = nCr + 2, then r =
13.
A. If 30C2r + 3 = 30C3r − 2 then r =
14.
15.
A. If 2nC3 : nC2 = 12:1 then what is n
B. If the ratio C(2n,3) : C(n,3) = 11:1 Find n.
C. If 2n C3 : n C2 = 10 :3 then n =
D. If 2n C3 : n C2 = 11 :1 then n =
16.
A. n + 2 C3n C3 =
B. n + 1 C3n − 1C3 =
17.
18.
19.
A. Prove by reasoning n + 1Cr = nCr + nCr − 1
B. Prove that n Cr = n − 1Cr − 1 + n − 1Cr
C. If n and r are positive integers such that r < n, then n Cr + nCr − 1 =
D. Find the value of nCr − 1 + nCr
E. Find the value of n − 1Cr − 1 + n − 1Cr
F. If n Cr + nCr + 1 = n + 1Cx then x =
G. If n and r are natural numbers such that 1 ≤ r ≤ n, prove that C(n,r) + C(n, r + 1) = C(n + 1, r + 1)
20.
A. Prove that 1 + C(3,1) + C(4,2) = C(5,3) B. Prove that C(2,1) + C(3,1) + C(4,1) = C(3,2) + C(4,2)
C. Verify 8C4 + 8C3 = 9C4 D. Find the value 10C4 + 10C5
E. Evaluate 25C2224C21 F. What is the value of 12C4 + 12C3?
21.
22.
23.
A. If n + 1C3 = 4 nC2 then n = B. If 10 nC2 = 3 n + 1C3, then n =
24.
A. If 15C4 + n + 5C5 = 16C5 then n =
25.
A. nCr + 2 nCr − 1 + nCr − 2=
B. Verify nCr + 2 nCr − 1 + nCr − 3= n + 2Cr
C. nCr + 1 + 2 nCr + nCr − 1=
D. Find the value of nCr + 2 nCr − 1 + nCr − 2
E. Prove that nCr = n − 2Cr − 2 + 2 n - 2Cr - 1 + n − 2Cr
26.
27.
28.
29.

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