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
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02. |
Solution » Hide/Show
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03. |
Find the value of 6C3
Solution » Hide/Show
6C3 |
= |
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[ |
nCr = |
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] |
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= |
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= |
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= = |
5 × 4 20 |
Alt
6C3 |
= |
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[ |
nCr = |
(n) (n − 1) ....... r times | r! |
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] |
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= |
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= = |
5 × 4 20 |
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04. |
Find n if nC2 = 36.
Solution » Hide/Show
nC2 = 36.
⇒ 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.
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05. |
nCr = nC n − r .
Solution » Hide/Show
L.H.S |
= |
nCr |
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= |
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R.H.S |
= |
nCn − r |
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= |
n! | (n −(n − r) )! (n − r)! |
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= |
n! | (n − n + r) )! (n − r)! |
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= |
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= |
L.H.S |
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06. |
Find the value of nCn − 3
Solution » Hide/Show
nCn − 3 |
= |
n! | (n −(n − 3) )! (n − 3)! |
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[ |
nCr = |
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] |
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= |
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= |
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n (n − 1) (n − 2) (n − 3)! | 3 × 2 × 1 × (n − 3)! |
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= |
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Alternative
nC(n − 3) |
= |
nC[n − (n − 3)] |
[nCr = nC(n − r)] |
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= |
nCn − n + 3 |
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= |
nC3 |
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= |
(n) × (n × 1) × (n × 2) | 3! |
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(n) × (n × 1) × (n × 2) | 3 × 2 × 1 |
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= |
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07. |
Find n if nCn − 4 = 15.
Solution » Hide/Show
nCn − 4 = 15.
⇒ |
n! | (n − (n − 4))! (n − 4)! |
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= 15 |
[ |
nCr = |
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] |
⇒ |
n (n − 1) (n − 2) (n − 3) (n − 4) | 4! × (n − 4)! |
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= 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! |
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= 15 |
[ |
nCr = |
n × (n − 1) × (n − 2) ... r times | r! |
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] |
⇒ 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
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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! |
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= |
9 × |
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⇒ |
(n + 1) × (n − 2) | 4 × 3 × 2 × 1 |
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= |
9 × |
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⇒ (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
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09. |
If nCr = 126 find n.
Solution » Hide/Show
nCr = 126
⇒ r = 0 → nC0 = 126 not relevant since nC0 = 1 |
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r = 1 → nC1 = 126 ⇒ |
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= 126 ⇒ n = 126 |
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r = 2 &rarr nC2 = 126 ⇒ |
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= 126 |
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⇒ (n) (n − 1) = 126 |
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⇒ 17 × 16 = 272; 16 × 15 = 240 ; → Not Possible |
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r = 3 &rarr nC3 = 126 ⇒ |
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= 126 |
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⇒ (n) (n − 1) (n − 2) = 126 × 3! |
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= 126 × 6 |
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r = 4 &rarr nC4 = 126 ⇒ |
(n) (n − 1) (n − 2) (n − 3) | 4! |
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= 126 |
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⇒ (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) |
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10. |
If nC6 = nC10 find n and evaluate nC4
Solution » Hide/Show
nC6 = nC10
⇒ 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 |
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= |
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16 × 15 × 14 × 13 | 4 × 3 × 2 × 1 |
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2 × 5 × 14 × 13 |
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= |
1,820 |
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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
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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
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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
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14. |
Find n if nC3 : n − 1C4 = 8 : 5
Solution » Hide/Show
nC3 : n − 1C4 = 8 : 5
⇒ |
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(n − 1) (n − 2) (n − 3) (n − 4) | 4! |
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= |
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⇒ |
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× |
4! | (n − 1) (n − 2) (n − 3) (n − 4) |
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= |
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⇒ |
(n) × 4 × 3! | (n − 3) (n − 4) × 3! |
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= |
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⇒ 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
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15. |
If 2nC3 : 2nC2 = 8 find n
Solution » Hide/Show
2nC3 : 2nC2 = 8
⇒ |
(2n) (2n − 1) (2n − 2) | 3 × 2 × 1 |
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× |
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= 8 |
⇒ 2 (n − 1) = 3 × 8
⇒ n − 1 = 12
⇒ n = 12 + 1
⇒ n = 13
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16. |
2nC2 − 2 nC2 =
Solution » Hide/Show
2nC2 − 2 nC2
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= |
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= |
[(2n) (2n − 1)] − 2 [(n) (n − 1)] | 2! |
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2n [(2n − 1) − (n − 1)] | 2 × 1! |
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n × n |
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n2 |
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17. |
If P = mC2 , find pC2
Solution » Hide/Show
P = mC2
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= |
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(m) (m − 1) [m2 − m − 2] |
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= |
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(m) (m − 1) [m2 − 2m + m − 2] |
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= |
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(m) (m − 1) [m (m − 2) + 1 (m − 2) ] |
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= |
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(m) (m − 1) [ (m + 1) (m − 2) ] |
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= |
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(m + 1) (m) (m − 1) (m − 2] |
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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
3n Cn + r |
= |
3n! | (3n − [n + r])! × (n + r)! |
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3n! | (3n − [n + n])! × (n + n)! |
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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 |
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= |
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+ |
n! | [n − ( r − 1)]! (r − 1)! |
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= |
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[L.C.M of r!, (r − 1)! is r!] [L.C.M of (n − r + 1)!, (n − r)! is (n − r + 1)!] |
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(n − r + 1) n! + (r) (n!) | (n − r + 1)! r! |
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n! [n − r + 1 + r ] | (n − r + 1)! r! |
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n! (n + 1) | [(n + 1) − r]! r! |
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(n + 1)! | [(n + 1) − r]! r! |
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= |
n + 1Cr |
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= |
R.H.S |
This is better remembered as n Cr − 1 + nCr = n + 1 Cr
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20. |
Prove that 20C13 + 20C14 = 20C6 + 20C7
Solution » Hide/Show
∴ L.H.S = R.H.S
Alternative I
20C13 + 20C14 |
=20C(20 − 13) + 20C(20 − 14) |
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=20C7 + 20C6 |
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= R.H.S |
Alternative II
LHS |
= |
20C13 + 20C14 |
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= |
21C14 |
[nC(r − 1) + nCr = (n + 1)Cr |
RHS |
= |
20C7 + 20C6 |
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= |
21C7 |
[ nCr + nC(r − 1) = (n + 1)Cr] |
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= |
21C(21 − 7) |
[nCr = nC(n − r)] |
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= |
21C14 |
⇒ L.H.S = R.H.S
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21. |
Evaluate 15C8 + 15C9 − 15C6 − 15C7
Solution » Hide/Show
15C8 + 15C9 − 15C6 − 15C7 |
= |
15C(15 − 8) + 15C(15 − 9) − (15C6 + 15C7) |
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= |
15C7 + 15C6 − (15C6 + 15C7) |
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(15C7 + 15C6) − (15C6 + 15C7) |
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= |
0 |
Alternative
15C8 + 15C9 − 15C6 − 15C7 |
= |
(15C9 + 15C8) − (15C7 + 15C6) |
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= |
16C9 − 16C7 |
[ nCr + nC(r − 1) = (n + 1)Cr] |
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= |
16C9 − 16C(16 − 7) |
[nCr = nC(n − r)] |
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= |
16C9 − 16C9 |
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= |
0 |
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22. |
If 8Cr − 7C3 = 7C2 Find r
Solution » Hide/Show
8Cr − 7C3 = 7C2
⇒ 8Cr = 7C2 + 7C3
⇒ 8Cr = 21 + 35
⇒ 8Cr = 56
⇒ 8Cr = 8 × 7
⇒ 8Cr |
= |
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⇒ 8Cr |
= |
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⇒ 8Cr |
= |
8C3 |
⇒ r = 3
Method II
8Cr − 7C3 = 7C2
⇒ 8Cr = 7C2 + 7C3
⇒ 8Cr = 8C3
[nC(r − 1) + nCr = (n + 1)Cr]
⇒ r = 3
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23. |
Find n if n + 1C3 = 2 nC2
Solution » Hide/Show
n + 1C3 = 2 nC2
⇒ |
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 nC2 − nC2
⇒ nC3 = nC2
⇒ n = 3 + 2
[nCa = nCb ⇒ n = a + b (Or) a = b]
⇒ n = 5
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24. |
If 8C3 + (n + 2)C4 = 9C4 then n =
Solution » Hide/Show
8C3 + (n + 2)C4 = 9C4
⇒ (n + 2)C4 = 9C4 − 8C3
(n + 2)C4 = 8C4 + 8C3 − 8C3
[(n + 1)Cr = nCr + nC(r − 1)]
⇒ (n + 2)C4 = 8C4
⇒ n + 2 = 8
⇒ n = 8 − 2
⇒ n = 6
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25. |
Prove that nCr + 2 nCr − 1 + nCr − 2 = n + 2Cr
Solution » Hide/Show
L.H.S |
= |
nCr + 2 nCr − 1 + nCr − 2 |
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= |
nCr + nCr − 1 + nCr − 1+ nCr − 2 |
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= |
(nCr + nCr − 1 ) + (nCr − 1+ nCr − 2) |
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= |
(n + 1Cr) + (n + 1Cr − 1) |
[nCr + nCr − 1 = n + 1Cr) |
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= |
([n + 1] + 1)Cr |
[nCr + nCr − 1 = n + 1Cr) |
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= |
(n + 1 + 1)Cr |
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= |
n + 2Cr |
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= |
R.H.S |
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26. |
Find the value of 8C4 + 2 8C3 + 8C2
Solution » Hide/Show
8C4 + 2 8C3 + 8C2
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= |
(8C4 + 8C3) + (8C3 + 8C2)
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= |
9C4 + 9C3
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[nCr + nCr − 1 = n + 1Cr) |
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= |
10C4
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[nCr + nCr − 1 = n + 1Cr) |
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= |
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= |
10 × 9 × 8 × 7 | 4 × 3 × 2 × 1 |
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= |
10 × 3 × 7 |
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= |
210 |
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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
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28. |
2nCn + 1 + 2 2nCn + 2nCn − 1 =
Solution » Hide/Show
2nCn + 1 + 2 2nCn + 2nCn − 1
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= |
2nCn + 1 + 2nCn + 2nCn + 2nCn − 1
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= |
(2n + 1)Cn + 1 + (2n + 1)Cn
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= |
((2n + 1) + 1)Cn + 1
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= |
(2n + 1 + 1)Cn |
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= |
(2n + 2)Cn |
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= |
2(n + 1)Cn |
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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
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= |
nCr + nCr − 1
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= |
n + 1Cr
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= |
RHS |
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No. |
Problems for Practice |
01. |
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02. |
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03. |
A. |
Find the value of 15C4
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B. |
How much is 21C19 =
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C. |
8C5 = ?
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D. |
Find the value of 9C0
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E. |
Evaluate 100C2
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F. |
What is the value of 7C1
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G. |
Find the value of 35C32
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H. |
Evaluate 14C14
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I. |
Find the value of 10C8
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04. |
A. |
If nC3 = 84 then n =
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B. |
Find n if nC2 = 45
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C. |
Find n ifnC4 = 126
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05. |
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06. |
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Prove that nC2 = |
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(n) (n − 1) |
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07. |
A. |
Find n in the following nCn − 3 = 10
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08. |
E. |
n + 2Cn = 45 find n
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F. |
4 nC2 = n + 2C3 find n
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09. |
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10. |
A. |
Find n if nC20 = nC35
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B. |
nC10 = nC15 find the value of 27Cn
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C. |
nC16 = nC9 find nC23 and 27Cn
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D. |
11Cx = 11Cyand x ≠y what is the value of x + y
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E. |
If C(n,8) = C(n,6) find C(n,2)
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F. |
n C10 = nC8, 21Cn =
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11. |
A. |
If 18Cr = 18Cr + 2, find the value of 'r' and rC3
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B. |
If C(10,r) = C(10, r + 4) Find the value of C(5, r)
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C. |
15C2r + 3 = 15C3r + 2 find 10Cr
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D. |
36C2r − 3 = 36C2r − 1 then r =
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E. |
If 35Cn + 7 = 35C4n − 2 then the values of n are given by ____.
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F. |
Find r in the following 15C(r + 3) = 15C(2r − 3)
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G. |
Find r in the following 20C3n = 20C2n + 5
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H. |
20C2r − 3 = 20C3r − 2 then r =
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12. |
A. |
If nCr = nCr + 2, then r =
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13. |
A. |
If 30C2r + 3 = 30C3r − 2 then r =
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14. |
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15. |
A. |
If 2nC3 : nC2 = 12:1 then what is n
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B. |
If the ratio C(2n,3) : C(n,3) = 11:1 Find n.
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C. |
If 2n C3 : n C2 = 10 :3 then n =
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D. |
If 2n C3 : n C2 = 11 :1 then n =
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16. |
A. |
n + 2 C3 − n C3 =
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B. |
n + 1 C3 − n − 1C3 =
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17. |
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18. |
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19. |
A. |
Prove by reasoning n + 1Cr = nCr + nCr − 1
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B. |
Prove that n Cr = n − 1Cr − 1 + n − 1Cr
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C. |
If n and r are positive integers such that r < n, then n Cr + nCr − 1 =
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D. |
Find the value of nCr − 1 + nCr
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E. |
Find the value of n − 1Cr − 1 + n − 1Cr
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F. |
If n Cr + nCr + 1 = n + 1Cx then x =
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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)
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20. |
A. |
Prove that 1 + C(3,1) + C(4,2) = C(5,3)
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B. |
Prove that C(2,1) + C(3,1) + C(4,1) = C(3,2) + C(4,2)
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C. |
Verify 8C4 + 8C3 = 9C4
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D. |
Find the value 10C4 + 10C5
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E. |
Evaluate 25C22 − 24C21
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F. |
What is the value of 12C4 + 12C3?
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21. |
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22. |
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23. |
A. |
If n + 1C3 = 4 nC2 then n =
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B. |
If 10 nC2 = 3 n + 1C3, then n =
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24. |
A. |
If 15C4 + n + 5C5 = 16C5 then n =
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25. |
A. |
nCr + 2 nCr − 1 + nCr − 2=
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B. |
Verify nCr + 2 nCr − 1 + nCr − 3= n + 2Cr
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C. |
nCr + 1 + 2 nCr + nCr − 1=
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D. |
Find the value of nCr + 2 nCr − 1 + nCr − 2
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E. |
Prove that nCr = n − 2Cr − 2 + 2 n - 2Cr - 1 + n − 2Cr
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26. |
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27. |
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28. |
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29. |
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