Factorials :: Solving equations involving factorials. Proving Complex relations

No. Problems & Solutions [Click Hide/Show to display the solutions below the question]
01.
If n! = 40320 then n =

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n! = 40320    
n! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8
n! = 8!
n = 8
02.
If (n +1)! = 6 (n - 1) find n.

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Given, (n + 1) = 6 (n − 1)!
(n + 1) (n) (n − 1)! = 6 (n − 1)!
(n + 1) (n) = 6
n2 + n = 6
n2 + n − 6 = 0
n2 + 3 n − 2n − 6 = 0
n (n + 3) − 2 (n + 3) = 0
(n − 2) (n + 3) = 0
(n − 2) = 0 (or) n + 3 = 0
n = 2 (or) n = − 3
n = 2

Alt
Given, (n + 1)! = 6 (n − 1)
(n + 1) (n) (n − 1)! = 6 (n − 1)
(n + 1) (n) = 6
(3) (2) = 6
on comparision n = 2.

03.
Prove that n (n - 1) (n - 2) ........... (n - r + 1) =
n!
(n - r)!

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R.H.S =
n!
(n − r)!
  =
(n) (n − 1) (n − 2) ------ (n − r + 1) (n − r)!
(n − r)!
  = (n) (n − 1) (n − 2) ------ (n − r + 1)
  = L.H.S
04.
1
2!
+
2
3!
+
3
4!
+ ... +
n
(n + 1)!
=

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05.
1 + 1 × 1! + 2 × 2! + 3 × 3! + ......... + n × n! =

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06.
1
3.1!
+
1
4.2!
+
1
5.3!
+ ... ∞ = ?

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07.
Prove that
n!
(n - r)! r!
+
n!
(r - 1)! (n - r + 1)!
=
(n + 1)!
r! (n - r + 1)!

Solution » Hide/Show

L.H.S =
n!
(n − r)! r!
+
n!
(r − 1)! (n − r + 1)!
[L.C.M of (n − r)! (n − r + 1)! is (n − r + 1)! ,
L.C.M of r!, (r − 1)! is r!]
  =
n! (n − r + 1) + n! r
(n − r + 1)! r!
  =
n! [n − r + 1 + r]
(n − r + 1)! r!
  =
n! (n + 1)
r! (n − r + 1)!
  = R.H.S
08.
Point out the mistake if any in the following reasoning
n!
(n - r)!
= (n) (n - 1) (n - 2) ...... (n - r + 1)

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09.
Prove that n! + 1 is not divisible by any number between 2 and n.

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n! = 1 × 2 × 3 × ------ × (n − 1) × n
Therefore n! is divisible by any number between 2 to n
n! + 1 = (1 × 2 × 3 × ------ × (n − 1) × n) + 1
Thus n! + 1 leaves a remainder 1 when divided by n!

Eg:-
5! = 5 × 4 × 3 × 2 × 1 (or) 1 × 2 × 3 × 4 × 5
  = 120

5! is divisible by any number between 2 and 5
5! + 1 = 120 + 1
  = 121
5! + 1
5
= 24 remainder 1
5! + 1
4
= 30 remainder 1
5! + 1
3
= 40 remainder 1
5! + 1
2
= 60 remainder 1
Thus 5! + 1 leaves a remainder 1 when divided by any number between 2 and 5.

10.
Show that 2n! = 2 n . n! [1.3.5.........(2n - 1)].

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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 − 3) (2n − 1)]
  = [2 × 2 × ----- n times) × [(n) (n − 1) (n − 2) ---- 3 × 2 × 1] × [1 × 3 × 5 × ----- (2n − 3) (2n − 1)]
  = 2n × n! × [1 . 3. 5 ---- (2n − 1)]
11.
Prove that 33! is divisible by 2 15 .

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33! = 33 × 32!
  = 33 × (2 × 16)!
  = 33 × 216 × 16! × [1. 3. 5. ----- × (2 × 16 − 1)]
  = 33 × 216 × 16! × [1. 3. 5. ----- × 31]
  = 215 [33 × 2 × 16! × (1. 3. 5. ----- × 31]
33! is divisible by 215  
12.
What is the largest integer 'n' such that 33! is divisible by 2 n .

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33! = 33 × 32!
  = 33 × (2 × 16)!
  = 33 × 216 × 16! × [1. 3. 5. ----- × (2 × 16 − 1)]
  = 33 × 216 × 16! × [1. 3. 5. ----- × 31]
  = 33 × 216 × (2 × 8)! × (1. 3. 5. ----- 31]
  = 33 × 2 16 × [28 × 8! × (1. 3. 5 ----(2 × 8 − 1)] × [(1. 3. 5 ---- 31)]
  = 33 × 2 16 × 28 × (2 × 4)! (1. 3. 5 ---- 15) (1. 3. 5 ---- 31)]
  = 33 × 2 24 × [24 × 4! × (1. 3. 5 ---- (2 × 4 − 1))] [(1. 3. 5 ---- 15) (1. 3. 5 ---- 31)]
  = 33 × 2 24 × 24 × (2 × 2)! (1. 3. 5 ---- 7) (1. 3. 5 ---- 15) (1. 3. 5 ---- 31)]
  = 33 × 2 28 × [22 × 2! × (1---- (2 × 2 − 1))] (1. 3. 5 ---- 7) (1. 3. 5 ---- 15) (1. 3. 5 ---- 31)]
  = 33 × 2 28 × 22 × 2 × (1.3) (1. 3. 5 ---- 7) (1. 3. 5 ---- 15) (1. 3. 5 ---- 31)]
  = 33 × 2 31 × (1.3) (1. 3. 5 ---- 7) (1. 3. 5 ---- 15) (1. 3. 5 ---- 31)]
33! is divisible by 231  
Maximum value of n for which 33! is divisible by 2n is 31.
No. Problems for Practice
02.
A. If (n + 1)! = 12 (n - 1)! find n.
10.
A. Show that 2n ! = 2 n (n!) [1.3.5............(2n - 1)] ∀ n ∈ N

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