# Finding amount of numbers.

• Mar 9th 2010, 10:49 PM
Monika1987
Finding amount of numbers.
I need help with finding how many 8-digit numbers there are containing no more than three zeros?
For exampel numbers like 10001111.

• Mar 9th 2010, 11:54 PM
Swlabr
Quote:

Originally Posted by Monika1987
I need help with finding how many 8-digit numbers there are containing no more than three zeros?
For exampel numbers like 10001111.

Can you think of a way of finding out how many numbers have no zeros? one zero? Two zeros? Three zeros?

(Hint: you have 8 places, and you are choosing none to be a zero, then one to be a zero, then two to be a zero, then three to be a zero...)

Finally, subtract these from the total number of numbers to get your result!
• Mar 10th 2010, 12:28 AM
chisigma
A quick alternative is the use of 'binomial coefficients'. The number of way to dispose k zeroes in a numer of n digits is ...

$\binom{n}{k} = \frac{n!}{k!\cdot (n-k)!}$ (1)

... so that the number of 8 digits numbers having no more than 3 zeroes is...

$N= \binom{8}{0} + \binom{8}{1} + \binom{8}{2}+ \binom{8}{3}$ (2)

Kind regards

$\chi$ $\sigma$