Combinations Notations :: Problems & Solutions

No. Problems & Solutions [Click Hide/Show to display the solutions below the question]
48.
Prove that C(2n, n) = 2n
1.3.5. .... (2n − 1)
n!

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C(2n, n) =
2n!
(2n − n)! n!
  =
2n!
n! n!
  =
2n n! [1. 3. 5 ----- (2n − 1)]
n! n!
  =
2n [1. 3. 5 ----- (2n − 1)]
n!
  = R.H.S

Hint:
2n! = (2n) (2n − 1) (2n − 2) (2n − 3) (2n − 4) -------- 6 × 5 × 4 × 3 × 2 × 1
  = [ (2n) (2n − 2) (2n − 4) ----- 6 × 4 × 2] [ (2n −1 ) (2n − 3) ------ 5 × 3 × 1]
  = [ [(2) (n)] [2 (n − 1)] [ 2 (n − 2)] ------ [2 (3)] [2 (2)] [2(1)] ] [1. 3. 5 ------ (2n − 1)]
  = (2 × 2 × 2 × ------ n times) [ (n) (n − 1) (n − 2) ------ 3 × 2 × 1] [1. 3. 5 ------ (2n − 1)]
  = 2n n! [1. 3. 5 ------ (2n − 1)]

50.
Prove 4nC2n : 2nCn = [1.3.5........(4n - 1)] : [1.3.5......(2n - 1 )] 2

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L.H.S = 4nC2n : 2nCn
  = 4nC2n ÷ 2nCn
  =
4n!
(4n − 2n)! 2n!
÷
2n!
(2n − n)! n!
  =
4n!
2n! 2n!
÷
2n!
n! n!
  =
4n!
2n! 2n!
×
n! n!
2n!
  =
4n! × (n!)2
(2n!)3
  =
22n 2n! × [1. .3 . 5 ----- (4n − 1)] × (n!)2
(2n!) (2n!)2
  =
( 2n)2 [1. 3. 5. ----- (4 n − 1)] ×
  =
( 2n)2 [1. 3. 5. ----- (4 n − 1)] ×
  =
( 2n)2 [1. 3. 5. ----- (4 n − 1)]
1
(2n)2 [1. .3 . 5 ----- (2n − 1)]2
  =
1. 3. 5 ------ (4n − 1)
[1. .3 . 5 ----- (2n − 1)]2

Hint:
4n! = (4n) (4n − 1) (4n − 2) (4n − 3) (4n − 4) --------------- 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
  = [(4n) (4n − 2) (4n − 4) -------- 8 × 6 × 4 × 2] [(4n − 1) (4n − 3) -------- 7 × 5 × 3 × 1]
  = [ [(2) (2n)] [2 (2n − 1)] [ 2 (2n − 2)] ------ [2 (4)] [2 (3)] [2 (2)] [2(1)] ] [1. 3. 5 ------ (4n − 1)]
  = (2 × 2 × 2 × ------ 2 n times) [ (2n) (2n − 1) (2n − 2) ------ 4 × 3 × 2 × 1] [1. 3. 5 ------ (4n − 1)]
  = 22n 2n! [1. 3. 5 ------ (4n − 1)]

Hint:
2n! = (2n) (2n − 1) (2n − 2) (2n − 3) (2n − 4) -------------- 6 × 5 × 4 × 3 × 2 × 1
  = [(2n) (2n − 2) (2n − 4) -------- 6 × 4 × 2] [(2n − 1) (2n − 3) -------- 5 × 3 × 1]
  = [ [(2) (n)] [2 (n − 1)] [ 2 (n − 2)] ------ [2 (3)] [2 (2)] [2(1)] ] [1. 3. 5 ------ (2n − 1)]
  = (2 × 2 × 2 × ------ n times) [ (n) (n − 1) (n − 2) ------ 3 × 2 × 1] [1. 3. 5 ------ (2n − 1)]
  = 2n n! [1. 3. 5 ------ (2n − 1)]

51.
Prove that the product of r consecutive positive integers is divisible by r!

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41.
Show that
(n) (n + 1) (n + 2) ------- (n + r − 1)
r!
is an integer.

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(n) (n + 1) (n + 2) ------- (n + r − 1)
r!
=
(n) (n + 1) (n + 2) ------- (n + r − 1)
r!
×
(n − 1)!
(n − 1)!
  =
(n − r − 1)!
(n − 1)! r!
  = n + r − 1Cr which is an integer

52.
Prove that 4C1 + 4C2 + 4C3 + 4C4 = 2 4 − 1

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53.
Show that nC0 + n + 1C1 + n + 2C2 +............... nCr = n + 1 C r + 1

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56.
9C5 + 9C4 . 6C1 + 9C3 . 6C2 + 9C2 . 6C3 + 9C1 . 6C4 + 6C5 =

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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

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22.
The value of n
Σ
r=1
npr
r!

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61.
Find the value of 47 C 4 + 5
Σ
r=1
(52 - r) C 3

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62.
Prove that n
Σ
k=m
k C r = n + 1 C r + 1m C r + 1

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63.
Evaluate
20Cr
25CR
When both numerator and denominator have their greatest values.

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20Cr n = 20, r = r
Since n = 20 is even,
  The term with the maximum value = The term whose value of r =
n
2
  = The term with r =
20
2
  = The term with r = 10
  ∴ Maximum value of 20Cr is 20C10

25CR n = 25, r = R
Since n = 20 is odd,
  The term with the maximum value = The term whose value of r is either =
n − 1
2
or
n + 1
2
  = The term with r =
20 − 11
2
or
25 + 1
2
  = The term with r =
24
2
or
26
2
  = The term with r = 12 or 13
  ∴ Maximum value of 25CR is 25C12 (or) 25C13

Greatest value of 20Cr
Greatest value of 25CR
=
20C10
25C12
  =
20!
(20 − 10)! 10!
(25)!
(25 − 12)! 12!
  =
20!
10! 10!
×
13! 12!
25!
  =
20! × 13 × 12 × 11 × 10! × 12 × 11 × 10!
10! × 10! × 25 × 24 × 23 × 22 × 21 × 20!
  =
13 × 12 × 11 × 12 × 11
25 × 24 × 23 × 22 × 21
  =
143
4025

64.
The difference between the greatest values of 15C r and 12C r is.

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65.
If y = 8C x, find what value of x gives a maximum value of y

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66.
Prove that the greatest value of 2nC r,is double the greatest value of 2n − 1Cr.

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67.
If a n = n
Σ
r=0
1
nCr
, then n
Σ
r=0
r
nCr
equal to

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68.
If C k for nCk , then n
Σ
k=1
is equal to
n (n + 1 ) (2n + 1)
24

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No. Problems for Practice
48.
2 n [1.3.5........ (2n − 1)]
n!
=
50.

A.
4nC2n
2nCn
=
B.
If 4nC2n :2nCn = k :[1.3.5..... (2n - 1 )] 2 then k =
56.
A. 6C4 + 6C3 . 5C1 + 6C2 . 5C2 + 6C1 . 5C3 + 5C4 =
61.

A.
The value of the expression 47C4 + 5
Σ
j=1
52jCj is equal to

B.
Show that 14C4 + 4
Σ
j=1
18 − jC3 = 18C4
63.
A. The ratio of 24Cr to 25Cr when each of them has their greatest value possible is.

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