Question

How many 4- digit numbers contain at most 2 zeroes? (Caution: A 4- digit number does not start with zero)

Answer 1

people if you can answer this question I would be very greatful

Answer 2

any ideas?

Answer 3

well the thing is it was on one of my tests. My lecturer doesn't have perfect English so she sometimes mixes up phrases and I need to see if she actually meant exactly 2 zeroes. I know the solution I did was in the right direction but I am pretty sure I left out something. My answer was 8334 including the numbers with 1 and 2 zeros.

Answer 4

Anyone??? help

Answer 5

to me :
with 1 zero :
| 9 | * | 9 | | 9 | |10| = 9^3 * 3!/2!

with 2 zeroes :
| 9 | * | 9 | |10| |10| = (90^2) * 3!/2!

Answer 6

oppsss..
with 1 zero :
| 9 | * | 9 | | 9 | |10| = 9^3 * 10 * 3!/2!

Answer 7

that can't be right because that would give me a 5 digit answer and there are only 9000 4-digit numbers

Answer 8

try again :)

no zeroes :
9.9.9.9 = 9^4 = 6561

1 zero :
9.9.9(0) = 9^3 = 729 * 3!/2! = 2187

2 zeroes :
9.9.(0).(0) = 9^2 * 3!/2! = 243
------------------------- +
total = 6561 + 2187 + 243 = 8991

hmmm --''

Answer 9

Thanks that looks better and seems right the way I was going......question? so when you looked at the question you immediately assumed they meant no zeros, 1 zero and 2 zeros right??? I have to ask because I think my lecturer wanted exactly two zeros but used the wrong phrase in the question

Answer 10

at most 2 means : there are 2, there is 1 and no ...

Answer 11

@satellite73
can you confirm this, if me right or no

Answer 12

yes that is exactly what I thought it's just that my lecturer does not have excellent English and she has been using wrong phrases for thing through the term and I have noticed it a lot. This question I had on my test and I went the same way about but after getting the 1 zero and 2 zeros I realized she more than likely meant something else and I stopped there and looking at how she was correcting it it seems I was right. She actually wanted to say exactly 2 zeroes

Answer 13

I'll talk to my lecturer just to see what she has to say. she knows all this just English is dragging her knowledge down

Answer 14

actually, my english so bad too. Dont call the police. Hehe :)

Answer 15

that's okay but that question might have caused my fail on the exam if she used the wrong phrase so I'll have to find out....y the way English is not my first language either

Answer 16

well let's count the number of \(4\)-digit numbers \(ABCD\), \(A\ne0\) such that \(3\) of the digits are \(0\). if \(3\) digits are \(0\) they must be \(B,C,D\) leaving \(A\) to be any of \(1,\dots,9\) so there are \(9\) possible ways for \(3\) zeros. also, it is impossible for all \(4\) digits to be \(0\) as we are given \(A\ne0\).

how many total \(4\)-digit numbers are there? well, precisely \(9\times10^3\), so there are \(9\times10^3-9=9(10^3-1)\) such numbers with at most \(2\) zeros

Answer 17

Thanks both answers have the same result. I would have done it the 1st way because for some reason I always complicate everything but the second answer is perfect and easy I would say

Answer 18

np I just figured it may be slightly easier