Problems 1-5 : Find the value of.
Problem 1 :
8!
Problem 2 :
Problem 3 :
Problem 4 :
Problem 5 :
Problem 6 :
Evaluate the above expression, when n = 7 and r = 5.
Problem 7 :
Simplify the above expression for any n with r = 3.
Problem 8 :
Solve for n :
Problem 9 :
Solve for n :
n! + (n - 1)! = 30
Problem 10 :
What is the unit digit of the sum 2! + 3! + 4! + ........ + 22!?
1. Answer :
8! = 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1
= 40320
2. Answer :
= 10 ⋅ 9 ⋅ 8
= 720
3. Answer :
= 4
4. Answer :
5. Answer :
= 9 ⋅ 8 ⋅ 7
= 504
6. Answer :
= 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3
= 2520
7. Answer :
8. Answer :
n! = 5!
n = 5
9. Answer :
n! + (n - 1)! = 30
n(n - 1)! + (n - 1)! = 30
(n - 1)!(n + 1) = 30
(n + 1)(n - 1)! = 5 ⋅ 6
(n + 1)(n - 1)! = 5 ⋅ (3 ⋅ 2 ⋅ 1)
(n + 1)(n - 1)! = 5 ⋅ 3!
Equate (n + 1) to 5.
n + 1 = 5
n = 4
10. Answer :
2! = 2 ⋅ 1 = 2
3! = 3 ⋅ 2 ⋅ 1 = 6
4! = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24
5! = 5 ⋅ 4! = 5 ⋅ 24 = 120
6! = 6 ⋅ 5! = 6 ⋅ 120 = 720
7! = 7 ⋅ 6! = 7 × 720 = 5040
8! = 8 ⋅ 7! = 8 ⋅ 5040 = 40320
From 5! onwards for all n!, the unit digit is zero and hence the contribution to the unit digit is through (2! + 3! + 4!) only.
That is,
2! + 3! + 4! = 2 + 6 + 24
= 32
Therefore the required unit digit is 2.
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