# Counting problems

• May 30th 2008, 02:34 PM
robocop_911
Counting problems

Q How many strings of four decimal digits

a) do not contain the same digit twice
my ans: 4 x 9^2
b) end with an even digit
my ans: 5 x 10^3
c) have exactly three digits that are 9s

my ans: 4 x 9

Thanks for the help!
• May 30th 2008, 02:38 PM
icemanfan
Quote:

Originally Posted by robocop_911

Q How many strings of four decimal digits

a) do not contain the same digit twice
my ans: 4 x 9^2
b) end with an even digit
my ans: 5 x 10^3
c) have exactly three digits that are 9s

my ans: 4 x 9

Thanks for the help!

a) do not contain the same digit twice:
$10 \cdot 9 \cdot 8 \cdot 7 = 5040$

c) have exactly three digits that are 9's: Your answer is correct.
• May 30th 2008, 02:43 PM
Plato
[QUOTE=robocop_911;151794]Q How many strings of four decimal digits
a) do not contain the same digit twice
my ans: 4 x 9^2 NO
b) end with an even digit
my ans: 5 x 10^3 YES
c) have exactly three digits that are 9s
my ans: 4 x 9 YES
Part (a) is a simple permutation: (10)(9)(8)(7). No repeated digits.
• May 30th 2008, 02:56 PM
robocop_911
[quote=Plato;151796]
Quote:

Originally Posted by robocop_911
Q How many strings of four decimal digits
a) do not contain the same digit twice
my ans: 4 x 9^2 NO
b) end with an even digit
my ans: 5 x 10^3 YES
c) have exactly three digits that are 9s
my ans: 4 x 9 YES
Part (a) is a simple permutation: (10)(9)(8)(7). No repeated digits.

Can you please explain why it won't be 4 x 9^2?
From what I think, there are 4 places for the same digit to occur twice in a 4 decimal digit and the remaining 2 places can contain any remaining 9 digits!
• Jun 12th 2008, 09:18 AM
robocop_911
Can anyone "check" my following question answer...
Q How many strings of four decimal digits

a) have exactly three digits that are 9s
my ans: 4 x 9
How come is it 4x9 can anyone explain me that?

Thanks for the help!
• Jun 12th 2008, 10:42 AM
Plato
There are 4 places to put the 'other' digit and 9 possible choices for that digit.
• Jun 12th 2008, 11:12 AM
robocop_911
Quote:

Originally Posted by Plato
There are 4 places to put the 'other' digit and 9 possible choices for that digit.

How come there are 4 places to put "other digits". We need to have "exactly" three 9s in the number!

Are you taking complement of the sentence or something?
• Jun 12th 2008, 12:03 PM
Plato
X999
9X99
99X9
999X
Four places.