No. |
Problems & Solutions [Click Hide/Show to display the solutions below the question] |
48. |
Solution » Hide/Show
C(2n, n) |
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2n n! [1. 3. 5 ----- (2n − 1)] | n! n! |
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2n [1. 3. 5 ----- (2n − 1)] | n! |
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R.H.S |
Hint:
2n! |
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(2n) (2n − 1) (2n − 2) (2n − 3) (2n − 4) -------- 6 × 5 × 4 × 3 × 2 × 1 |
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[ (2n) (2n − 2) (2n − 4) ----- 6 × 4 × 2] [ (2n −1 ) (2n − 3) ------ 5 × 3 × 1] |
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[ [(2) (n)] [2 (n − 1)] [ 2 (n − 2)] ------ [2 (3)] [2 (2)] [2(1)] ] [1. 3. 5 ------ (2n − 1)] |
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(2 × 2 × 2 × ------ n times) [ (n) (n − 1) (n − 2) ------ 3 × 2 × 1] [1. 3. 5 ------ (2n − 1)] |
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2n n! [1. 3. 5 ------ (2n − 1)] |
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50. |
Prove 4nC2n : 2nCn = [1.3.5........(4n - 1)] : [1.3.5......(2n - 1 )] 2
Solution » Hide/Show
L.H.S |
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4nC2n : 2nCn |
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4nC2n ÷ 2nCn |
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22n 2n! × [1. .3 . 5 ----- (4n − 1)] × (n!)2 | (2n!) (2n!)2 |
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( 2n)2 [1. 3. 5. ----- (4 n − 1)] |
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( 2n)2 [1. 3. 5. ----- (4 n − 1)] |
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( 2n)2 [1. 3. 5. ----- (4 n − 1)] |
1 | (2n)2 [1. .3 . 5 ----- (2n − 1)]2 |
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1. 3. 5 ------ (4n − 1) | [1. .3 . 5 ----- (2n − 1)]2 |
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Hint:
4n! |
= (4n) (4n − 1) (4n − 2) (4n − 3) (4n − 4) --------------- 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 |
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= [(4n) (4n − 2) (4n − 4) -------- 8 × 6 × 4 × 2] [(4n − 1) (4n − 3) -------- 7 × 5 × 3 × 1] |
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= [ [(2) (2n)] [2 (2n − 1)] [ 2 (2n − 2)] ------ [2 (4)] [2 (3)] [2 (2)] [2(1)] ] [1. 3. 5 ------ (4n − 1)] |
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= (2 × 2 × 2 × ------ 2 n times) [ (2n) (2n − 1) (2n − 2) ------ 4 × 3 × 2 × 1] [1. 3. 5 ------ (4n − 1)] |
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= 22n 2n! [1. 3. 5 ------ (4n − 1)] |
Hint:
2n! |
= (2n) (2n − 1) (2n − 2) (2n − 3) (2n − 4) -------------- 6 × 5 × 4 × 3 × 2 × 1 |
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= [(2n) (2n − 2) (2n − 4) -------- 6 × 4 × 2] [(2n − 1) (2n − 3) -------- 5 × 3 × 1] |
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= [ [(2) (n)] [2 (n − 1)] [ 2 (n − 2)] ------ [2 (3)] [2 (2)] [2(1)] ] [1. 3. 5 ------ (2n − 1)] |
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= (2 × 2 × 2 × ------ n times) [ (n) (n − 1) (n − 2) ------ 3 × 2 × 1] [1. 3. 5 ------ (2n − 1)] |
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= 2n n! [1. 3. 5 ------ (2n − 1)] |
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51. |
Prove that the product of r consecutive positive integers is divisible by r!
Solution » Hide/Show
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41. |
Show that |
(n) (n + 1) (n + 2) ------- (n + r − 1) | r! |
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is an integer. |
Solution » Hide/Show
(n) (n + 1) (n + 2) ------- (n + r − 1) | r! |
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(n) (n + 1) (n + 2) ------- (n + r − 1) | r! |
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n + r − 1Cr which is an integer |
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52. |
Prove that 4C1 + 4C2 + 4C3 + 4C4 = 2 4 − 1
Solution » Hide/Show
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53. |
Show that nC0 + n + 1C1 + n + 2C2 +............... nCr = n + 1 C r + 1
Solution » Hide/Show
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56. |
9C5 + 9C4 . 6C1 + 9C3 . 6C2 + 9C2 . 6C3 + 9C1 . 6C4 + 6C5 =
Solution » Hide/Show
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58. |
If nCn − r + 3 nCn − r + 1 + 3 nCn − r + 2 + nCn − r + 3 = pC r then prove that p = n + 3
Solution » Hide/Show
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22. |
Solution » Hide/Show
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61. |
Find the value of 47 C 4 + |
5 Σ r=1 |
(52 - r) C 3 |
Solution » Hide/Show
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62. |
Prove that |
n Σ k=m |
k C r = n + 1 C r + 1 − m C r + 1 |
Solution » Hide/Show
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63. |
Evaluate |
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When both numerator and denominator have their greatest values. |
Solution » Hide/Show
20Cr ⇒ |
n = 20, r = r
Since n = 20 is even, |
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The term with the maximum value |
= The term whose value of r = |
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= The term with r = |
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= The term with r = |
10 |
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∴ Maximum value of 20Cr is 20C10 |
25CR ⇒ |
n = 25, r = R
Since n = 20 is odd, |
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The term with the maximum value |
= The term whose value of r is either = |
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or |
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= The term with r = |
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or |
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= The term with r = |
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or |
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= The term with r = |
12 |
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13 |
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∴ Maximum value of 25CR is 25C12 (or) 25C13 |
Greatest value of 20Cr | Greatest value of 25CR |
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20! × 13 × 12 × 11 × 10! × 12 × 11 × 10! | 10! × 10! × 25 × 24 × 23 × 22 × 21 × 20! |
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13 × 12 × 11 × 12 × 11 | 25 × 24 × 23 × 22 × 21 |
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64. |
The difference between the greatest values of 15C r and 12C r is.
Solution » Hide/Show
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65. |
If y = 8C x, find what value of x gives a maximum value of y
Solution » Hide/Show
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66. |
Prove that the greatest value of 2nC r,is double the greatest value of 2n − 1Cr.
Solution » Hide/Show
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67. |
If a n = |
n Σ r=0 |
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, then |
n Σ r=0 |
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equal to |
Solution » Hide/Show
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68. |
If C k for nCk , then |
n Σ k=1 |
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is equal to |
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Solution » Hide/Show
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